OP Malhotra Class 12 Maths Solutions Chapter 17 Differential Equations Ex 17(g)

Well-structured ISC Class 12 OP Malhotra Solutions Chapter 17 Differential Equations Ex 17(g) facilitate a deeper understanding of mathematical principles.

S Chand Class 12 ICSE Maths Solutions Chapter 17 Differential Equations Ex 17(g)

Question 1.
Obtain the equation of the curve whose slope at any point is equal to y + 2x and which passes through the origin.
Solution:
We know that, slope of tangent at any point P (x, y) on the curve = \(\frac { dy }{ dx }\)
also given slope of tangent to the curve = y + 2x
∴ \(\frac { dy }{ dx }\) = y + 2x
⇒ \(\frac { dy }{ dx }\) – y = 2x
which in L.D.E in y of first order and is of dy the form \(\frac { dy }{ dx }\) + Py = Q
Here P = – 1 ; Q = 2x

Since given curve passes through origin i.e. x = Q,y = Q
∴ from (1); 0 = -2 (0 + 1) + c ⇒ c = 2
Thus from (1); y = – 2 (x + 1) + 2e^x
which is the required eqn. of curve.

Question 2.
The surface area of a balloon being inflated changes at a constant rate. If initially, its radius is 3 units and after 2 seconds, it is 5 units, find the radius after t second.
Solution:
Let r (t) be the radius of balloon after time t seconds
Let S be the surface area of balloon d/S
Then \(\frac { ds }{ dt }\) = k where k be the constant of proportionality

which is the required radius after time t.

Question 3.
A population grows at the rate of 8% per year. How long does it take for the population to double? Use differential equation for it.
Solution:
Let the initial population be P₀ and P be the population after t years.

Question 4.
The slope of a tangenrat point P (x, y) on the curve is – \(\frac { x }{ y }\). If the curve passes through the point (-3, 4), find the equation of the curve.
Solution:
We know that, slope of the tangent to the curve at any point P(x, y) = \(\frac { dy }{ dx }\)
also, given slope of tangent to given curve = – \(\frac { x }{ y }\)
∴ \(\frac{d y}{d x}=-\frac{x}{y}\) ⇒ y dy + x dx = 0
On integrating ; we have
∫ ydy + ∫ xdx = \(\frac { c }{ 2 }\)
⇒ \(\frac{y^2}{2}+\frac{x^2}{2}=\frac{c}{2}\)
⇒ x² + y² + c …(1)
Since the curve given by (1) passes through
the point (- 3, 4)
from (1) ; 9 + 16 = c ⇒ c = 25
Thus eqn. (1) becomes ; x² + y² = 25 be the required eqn. of curve.

Question 5.
The slope of the tangent to a curve at any point is reciprocal of twice the ordinate of that point. The curve passes through (4,3). Formulate the differential equation and hence find the equation of curve.
Solution:
We know that, slope of the tangent to a curve at any point P (x, y) = \(\frac { dy }{ dx }\)
also given slope of the tangent to curve = \(\frac { 1 }{ 2y }\)
∴ \(\frac { dy }{ dx }\) = \(\frac { 1 }{ 2y }\) ⇒ 2y dy = dx
On integrating ; we have
∫ 2ydy =∫ dx ⇒ y² = x + c …(1)
Since the curve given by (1) passes through point (4, 3)
∴ from (1); 9 = 4 + c ⇒ e = 5
∴ from (1) ; y² = x + 5
which is the required eqn. of curve.

Question 6.
The slope of tangent at any point to a curve is λ. times the slope of the line joining the point of contact to the origin. Find the equation of curve.
Solution:
It is given that, slope of tangent at any point (x, y) to a curve is A times the slope of the line joining the point of contact to the (0, 0).

Question 7.
The slope of the tangent to a curve at a point (x, y) on it is given by
\(\frac { y }{ x }\) – cot\(\frac { y }{ x }\).cos\(\frac { y }{ x }\)(x > 0, y > 0)
and curve passes through the point (1, \(\frac { π }{ 4 }\)). Find the equation of the curve.
Solution:
Since the slope of the tangent to a curve at a point is \(\frac { y }{ x }\) – cot\(\frac { y }{ x }\) cos\(\frac { y }{ x }\).

be the required eqn. of curve.

Question 8.
(i) Find the equation of the curve for which the intercept cut off by a tangent on the xvaxis is equal to four times the ordinate of the point of contact.
(ii) Similar question. The x-intercept of the tangent line to a curve is equal to the ordinate of the point of contact. Find the particular curve through the point (1,1).
Solution:
(i) eqn. of any tangent at any point (x, y) is given by
Y – y = \(\frac { dy }{ dx }\) (X – x) …(1)
For x-intercept, Y = 0 ∴ from (1); we have

be the required eqn. of curve.

(ii) The eqn. of tangent to curve at any point (x, y) is given by

which is the required eqn. of curve.

Question 9.
Find the differential equation of ail circles which pass through the origin and whose centre lies on the y-axis.
Solution:
eqn. of circle having centre on y-axis i.e. (0, b) and radius r is given by
(x – 0)² + (y – b)² = r² … (1)
since circle passes through origin
∴ 0² + (0 – b)² = r² ⇒ r² = b²
∴ eqn. (1) becomes ; x² + (y – b)² = b²
⇒ x² + y² – 2by = 0 … (2)
where b be the arbitrary constant
Diff. both sides of eqn. (2) w.r.t. x ;

which is the required eqn. of differential eqn.

Question 10.
The line normal to a given curve at each point (x, y) on the curve passes through the point (2, 0). If the curve contains the point (2,3) find its equation.
Solution:
The eqn. of normal to curve at any point (x, y) is given by
Y – y = – \(\frac{1}{\frac{d y}{d x}}(\mathrm{X}-x)\) … (1)
Clearly it is given that line (1) passes through the point (2, 0).
∴ 0 – y = – \(\frac{1}{\frac{d y}{d x}}(2-x)\)
⇒ y\(\frac { dy }{ dx }\) = 2 – x ⇒ y dy = (2 – x) dx
On integrating ; we have
\(\frac{y^2}{2}=\frac{(2-x)^2}{-2}+\frac{c}{2}\)
⇒ y² = – (2 – x)² + c … (2)
Since the curve given by eqn. (2) passes through (2, 3).
∴ from (2); we get
3² = – (2 – 2)² + c ⇒ c = 9
∴ from (2); y² = – (2 – x)² + 9 = 4x – x² + 5
which is the required eqn. of curve.

Question 11.
Assume that a radioactive substance decomposes at a rate proportional to the quantity of the substance present. In an experiment, with Radium 226, it was observed that in 25 years only 1.1 per cent of the quantity was decomposed. Find how long will take for one half of the original amount to decompose.
Solution:
Let P be the amount of radioactivesubstance present at any time t. Then
\(\frac { dP }{ dt }\) ∝P ⇒ \(\frac { dP }{ dt }\) = – KP
where K be the constant of proportionality

Question 12.
A certain radioactive material has a half life of 2 hours. Find the time interval required for a given amount of this material to decay to one tenth of its original mass.
Solution:
Let M be the quantity of mass at any time t
Then \(\frac { dM }{ dt }\) ∝M ⇒ \(\frac { dM }{ dt }\) = – KM
where K = constant of proportionality

also it is given that, radiactive material has a half life be 2 hours
∴ When t = 2 hours,
M = \(\frac{\mathrm{M}_0}{2}\)
∴ from (2); log \(\frac { 1 }{ 2 }\) = – K x 2

Question 13.
Experiments show that radium disinte-grates at a rate proportional to the amount of radium-present at the moment. Its half life is 1590 years. What percentage will disappear in one year?
[Use \(e^{\frac{\log 2}{1590}}\) = 0.99%]
Solution:
Let P the amount of radium present at any time t then
\(\frac { dP }{ dt }\) ∝P ⇒ \(\frac { dP }{ dt }\) = – KP
where K = constant of proportionality
\(\frac { dP }{ dt }\) = – K dt ; on integrating
log P = – Kt + c …(1)
Let P₀ be the amount of radium present initially
∴ t = 0, P = P₀
∴ from (1) ; we have log P₀ = c
from (1); log P = – Kt + log P₀
⇒ \(\log \frac{P}{P_0}\) = – Kt … (2)
Given half life of radioactive substance be 1590 years

Thus, the required % of radius will disappears in one year
= (100 – 99.96)% = 0.04%

Question 14.
A spherical raindrop evaporates at a rate proportional to its surface area. If originally its radius be 3 mm and 1 hour later it reduces to 2 mm, find an expression for the radius of the raindrop at any time.
Solution:
Let r (t) be the radius of raindrop after time t hours. Since the radius is decreases as t increases.
∴ the rate of change of r must be negative. Let V be the volume of the raindrop and S be the surface area of raindrop.
Also it is given that \(\frac { dV }{ dt }\) ∝ S
⇒ \(\frac{d}{d t}\left(\frac{4}{3} \pi r^3\right)=-\mathrm{K} \times 4 \pi r^2\)
where K = constant of proportionality
⇒ \(\frac{4}{3} \pi \times 3 r^2 \frac{d r}{d t}=-4 \pi r^2 \mathrm{~K}\)
⇒ \(\frac{d r}{d t}=-\mathrm{K} \Rightarrow d r=-\mathrm{K} d t\)
On integrating ; we have
r = – Kt + c …(1)
given t = 0, r = 3 mm ∴ from (1); c = 3 ,
∴ eqn. (1) becomes ; r = – Kt + 3 …(2)
when t = 1 hr; r = 2 mm
∴ from(2); 2 = – K + 3 ⇒ K = 1
Thus eqn. (2) becomes ; r = – t + 3
be the required radius after time t.
Aliter : Let r (t) be the radius of raindrop after time t hours. Since the radius r decreases as t increases. ∴ the rate of change of r must be negative.
Then \(\frac { dr }{ dt }\) ∝ S,
where S = surface area of raindrop

be the required radius after time t.

Question 15.
An equation relating to stability of an aeroplane is given by \(\frac { dv }{ dt }\) = g cos α – kv, where v is the velocity and g, α, k are constants. Find the expression for the velocity, if v = 0 at t = 0.
Solution:
Given eqn. related to stability of an aeroplane is given by
\(\frac { dv }{ dt }\) = g cos α – kv
⇒ \(\frac { dv }{ dt }\) + kv = g cos α
which is of the form \(\frac { dv }{ dt }\) + Pv = Q
where P = k; Q = g cos α

which is the expression for velocity after time t.

Question 16.
The acceleration of a particle moving in a straight line is (10 – 6t) cm/s² after t seconds. If the velocity of the particle is zero at t = 1/3, show that it will be zero at t = 3.
Solution:
Let a (t) be the acceleration of particle moving in a straight line after time t
Then a (t) = 10 – 6t ⇒ \(\frac { dv }{ dt }\) = 10 – 6t
⇒ dv = (10 – 6t) dt; on integrating
∫ dv = ∫ (10 – 6t)dt
⇒ v = 10t – 3t² + c … (1)
It is given that v = 0 at t = \(\frac { 1 }{ 3 }\)
∴ from (1); 0 = \(\frac { 10 }{ 3 }\) – 3 x \(\frac { 1 }{ 9 }\) + c ⇒ c = – 3
∴ from (1); we have
v = 10t – 3t² – 3
at t = 3; v = 10 x 3 – 3 x 3² – 3 = 0
Hence the velocity of the particle is zero at t = 3.

Question 17.
The population of a country doubles in 40 years. Assuming that the rate of increase is proportional to the number of inhabitants, find the number of years in which it will treble itself.
Solution:
Let P be the population after time t
Also it is given that \(\frac { dP }{ dt }\) ∝ P
⇒ \(\frac { dP }{ dt }\) = kP
where k = constant of proportionality
⇒ \(\frac { dP }{ P }\) = kdt; on integrating ; we have
log P = kt + c …(1)
Let P₀ be the population initially i. e. at t = 0
∴ log p₀ = c
∴ from (1); log P = kt + log P₀
⇒ \(\log \frac{P}{P_0}\) = kt …(2)
Given P = 2P₀ When t = 40 years

Thus after \(\frac{40 \log 3}{\log 2}\) years, population will become treble itself.

Question 18.
The rate of increase of bacteria in a culture is proportional to the number of bacteria present and it is found that the number doubles in 5 hours. Express this mathematically using rate of increase of bacteria with respect to time. Hence, calculate how many times the bacteria may be expected to grow at the end of 15 hours.
Solution:
Let x be the no. of bacteria after time t
Then \(\frac{d x}{d t} \propto x \Rightarrow \frac{d x}{d t}=k x\)
where k = constant of proportionality variable separation ; we have
\(\frac { dx }{ x }\) = kdt; on integrating ;
\(\int \frac{d x}{x}=\int k d t\)
⇒ log x = kt + c …(1)
Let x₀ be the population initially i. e. at t = 0
∴ from (1); log x₀ = c
Thus eqn. (1) becomes

Thus the required no. of bacteria may be expected to grow at the end of 15 years be eight times the no. of bacteria present initially.

Question 19.
A slow motorist levelling 1 m/s disengages gear and free wheels to rest. The retardation the car has two components 0.04 m/s2 due to friction in working parts and road resistance i retardation due to air resistance of 0.04 v² m/s², where v is the speed in m/s. Find how would it take the car to free wheel to rest?
Solution:
Let a be the retardation of the car
Then a = – 0.04 – 0.04 v²

∴ from (1); tan^-1 1 = c ⇒ c = \(\frac { π }{ 4 }\)
Thus eqn. (1) becomes ;
tan^-1 v = – 0.04 t + π/4 …(2)
The car goes to rest at t = t₁ sec.
∴ v = 0 at t = t₁
∴ from (2) ; tan^-1 0 = – 0.04 x t₁ + π/4
⇒ 0.04 t₁ = π/4 ⇒ t₁ = \(\frac { 25π }{ 4 }\) seconds

Question 20.
A wet porous substance in the open air loses its moisture at rate proportional to the moisture content. If a sheet hung in the wind loses half of its moisture during the first hour, when it have lost.
(i) 90% moisture, weather conditions remaining the same.
(ii) 95% moisture, weather conditions remaining the same.
(iii) 98% moisture, weather conditions remaining the same.
Solution:
Let M be the moisture content at time t
Then \(\frac{d \mathrm{M}}{d t} \propto \mathrm{M} \Rightarrow \frac{d \mathrm{M}}{d t}=-k \mathrm{M}\)
where k = constant of proportionality
Since moisture content loses after time t
∴ \(\frac { dM }{ dt }\) = – kdt; On integrating
\(\int \frac{d \mathrm{M}}{\mathrm{M}}=-k \int d t\)
⇒ log M = – kt + c …(1)
Let M₀ be the moisture content in the net porous substance initially i.e. at t = 0, M = M₀.
∴ from (1); log M₀ = c
Thus eqn. (1) becomes ;

Question 21.
The engine of a motor boat moving at 10 m/s is shut off. Given that the retardation at subsequent time (after shutting of the engine) equals the velocity at that time, find
(i) the velocity after 2 sees of switching off the engine;
(ii) the distance travelled in these 2 seconds (Leave your answer in terms of e).
Solution:
Let a be the retardation of engine of motor boat and v be the velocity of engine of motor boat.
Then a = – v ⇒ \(\frac { dv }{ dt }\) = – v ⇒ \(\frac { dv }{ v }\) = – dt
On integrating ; we have
\(\int \frac{d v}{v}=-\int d t\) ⇒ log v = – t + c …(1)
Given at t = 0 ; v = 10 m/s
∴ from (1); log 10 = c
Thus eqn, (1) becomes ;
\(\log \frac{v}{10}=-t \Rightarrow \frac{v}{10}=e^{-t}\)
⇒ v = 10 e^-t … (2)
(i) When t = 2 ;
velocity after 2 secs = 10 e^-2 m/sec

(ii) Let x be the distance travelled by motor boat
∴ \(\frac { dx }{ dt }\) = 10e^-t ⇒ dx = 10e^-t dt
On integrating ; we have
x = \(\frac{10 e^{-t}}{-1}\) + c … (3)
When t = 0 ; x = 0 from (3); we have
0 = – 10 + c ⇒ c = 10
∴ from (3); x = – 10 e^-t + 10
⇒ x = 10 (1 – e^-t)
Thus required distance travelled by motor boat in 2 seconds = (x)_t=2 = 10(1 – e^-2)

Question 22.
A steam boat is moving at velocity v0 when steam is shut off. Given that the retardation at any subsequent time is equal to the magnitude of the velocity at the time, find the velocity and distance travelled in time t after the steam is shut off.
Solution:
Given retardation at any subsequent time is equal to magnitude of the velocity at that time.
Let a be the retardation and v be the velocity after time t.
Then a = \(\frac { dv }{ dt }\) = – v ⇒ \(\frac { dv }{ dv }\) = – dt
On integrating ; we have
log v = – t + c … (1)
at t = 0 ; v = v₀
∴ from (1); log v₀ = c
Thus eqn. (1) becomes ; log \(\frac{v}{v_0}\) = – t
⇒ \(\frac{v}{v_0}=e^{-t} \Rightarrow v=v_0 e^{-t}\) … (2)
Let x be the distance travelled by steam boat after time t
∴ v = \(\frac { dx }{ dt }\) = v₀e^-t
On integrating x = v₀\(\frac{e^{-t}}{-1}\) + c₁ …(3)
When t = 0 ; x = 0
∴ from (3); 0 = – v₀ + c₁ ⇒ c₁ = v₀
Thus eqn. (3) becomes ; x = – v₀ e^-t + v₀
⇒ x = v₀ (1 – e^-t) be the required distance covered by steam boat.

Example

Solve the following differential equations:

Question 1.
x\(\frac { dy }{ dx }\) + y = 3x² – 2
Solution:
Given x\(\frac { dy }{ dx }\) + y = 3x² – 2 it can be written
as \(\frac{d y}{d x}+\frac{y}{x}=\frac{3 x^2-2}{x}=3 x-\frac{2}{x}\)
which is L.D.E in y of first order and is of the form \(\frac { dy }{ dx }\) + Py = Q
where P = \(\frac { 1 }{ x }\) and Q = 3x – \(\frac { 2 }{ x }\)
∴ I.F. = \(e^{\int \mathrm{P} d x}=e^{\int \frac{1}{x} d x}=e^{\log x}\) = x
and solution is given by
\(y \cdot e^{\int \mathrm{P} d x}=\int \mathrm{Q} \cdot e^{\int \mathrm{P} d x}\) dx + x
⇒ \(y \cdot x=\int\left(3 x-\frac{2}{x}\right)\) x dx + c
⇒ xy = x³ – 2x + c
which is the required solution.

Question 2.
\(\frac { dy }{ dx }\) – \(\frac { y }{ x }\) = x² – 2
Solution:
Given diff. eqn. \(\frac { dy }{ dx }\) – \(\frac { y }{ x }\) = x² – 2, which is L.D.E in y of first order and is of the form
\(\frac { dy }{ dx }\) + Py = Q ;

which is the required solution.

Question 3.
\(\frac { dy }{ dx }\) = e^x-y + x²^e-y
Solution:

Question 4.
\(\frac{d y}{d x}=\frac{(1+x) y^2}{x^2(y-1)}\)
Solution:

Question 5.
(1 – x²)\(\frac { dy }{ dx }\) + xy = x(1 – x²)^1/2
Solution:

Question 6.
\(\frac{d y}{d x}=e^{3 x-2 y}+x^2 e^{-2 y}\)
Solution:

Question 7.
\(\frac{1}{\sin ^{-1} x} \cdot \frac{d y}{d x}\) = 1
Solution:

Question 8.
(x + 1)\(\frac { dy }{ dx }\) – y = e^3x(x + 1)²
Solution:
Given diff. eqn. can be written as,
\(\frac{d y}{d x}-\frac{y}{x+1}=e^{3 x}(x+1)\)
which is L.D.E in y of first order and is of the form \(\frac { dy }{ dx }\) + Py = Q

which gives the required solution.

Question 9.
(x + 2y²)\(\frac { dy }{ dx }\) = y where y > 0
Solution:

Question 10.
y – x\(\frac { dy }{ dx }\) = a(y² + \(\frac { dy }{ dx }\))
Solution:

Question 11.
\(\frac { dy }{ dx }\) = x²(x – 2), given v = 2 when x = 0
Solution:
Given \(\frac { dy }{ dx }\) = x² (x – 2)

be the required solution.

Question 12.
\(\frac { dy }{ dx }\) + \(\frac { y }{ x }\) = x², given y – 1 when x = 1
Solution:
Given \(\frac { dy }{ dx }\) + \(\frac { y }{ x }\) = x²
which is linear in x of first order and is of the form \(\frac { dy }{ dx }\) + Py = Q.
Here P = \(\frac { 1 }{ x }\) ; Q = x²

be the required solution.

Question 13.
(1 + y²)dx = (tan^-1 y – x)dy
Solution:
Given, (1 + y²)dx = (tan^-1 y – x)dy

which is the required solution.

Question 14.
\(\frac { dy }{ dx }\) = (4x + y + 1)²
Solution:
Given \(\frac { dy }{ dx }\) = (4x + y + 1)²
put 4x + y + 1 = t; DifF. w.r.t. x both sides, we have
⇒ 4 + \(\frac { dy }{ dx }\) = \(\frac { dt }{ dx }\)

which gives the required solution.

Question 15.
\(\frac { dy }{ dx }\) = x(2logx + 1), given that when y = 0, x = 2.
Solution:
Given \(\frac { dy }{ dx }\) = x (2 log x + 1) dx

given when y = 0, x = 2
∴ from eqn. (1) ; we have
0 = 4 log 2 + c ⇒ c = – 4 log 2
Thus eqn. (1) becomes ;
y = x² log x – 4 log 2
be the required particular solution.

Question 16.
3e^x tan y dx + (1 – e^x)scc² y dy = 0
Solution:
Given 3e^x tan y dx + (1 – e^x)scc² y dy = 0
On dividing throughout by (1 – e^x) tan y ; we get

which gives the required solution.

Question 17.
\(\frac{d y}{d x}=\frac{y}{x}+\tan \left(\frac{y}{x}\right)\)
Solution:

Question 18.
\(\frac { dy }{ dx }\) + sin (x + y) = sin (x – y)
Solution:
Given \(\frac { dy }{ dx }\) + sin (x + y) = sin (x – y)
⇒ \(\frac { dy }{ dx }\) = sin (x – y) – sin (x + y) = 2 cos x sin (- y)
⇒ \(\frac { dy }{ sin y }\) = – 2 cos x dx
⇒ cosec y dy= – 2 cosx dx
On integrating both sides ; we have
∫ cosec y dy + ∫ 2 cos x dx = 0
⇒ log tan\(\frac { y }{ 2 }\) + 2sinx = c
which gives the required solution.

Question 19.
sin x\(\frac { dy }{ dx }\) – y = cos² x sin x tan \(\frac { x }{ 2 }\)
Solution:
Given diff. eqn. can be written as
\(\frac { dy }{ dx }\) – (cosec x) y = cos² x tan \(\frac { x }{ 2 }\)
which is L.D.E in y of first order and is of the form \(\frac { dy }{ dx }\) + Py = Q.

be the required particular solution.

Question 20.
x (x – y) dy + y²dx = 0
Solution:
Given diff. eqn. can be written as ;

which gives the required solution.

Question 21.
\(\frac{d y}{d x}+\frac{2 x}{1+x^2} y=\frac{1}{\left(1+x^2\right)^2}\)
Solution:
Given \(\frac{d y}{d x}+\frac{2 x}{1+x^2} y=\frac{1}{\left(1+x^2\right)^2}\)
which is linear in y of first order and is of the form \(\frac { dy }{ dx }\) + Py = Q

gives the required solution.

Question 22.
x\(\frac { dy }{ dx }\) – y = \(\sqrt{x^2+y^2}\)
Solution:
Given x\(\frac { dy }{ dx }\) – y = \(\sqrt{x^2+y^2}\)

which gives the required solution.

Question 23.
e^y(1 + y²)\(\frac { dy }{ dx }\) – \(\frac { x }{ y }\) = 0
Solution:

Question 24.
cos² x\(\frac { dy }{ dx }\) + y = tan x
Solution:
Given cos² x \(\frac { dy }{ dx }\) + y = tanx dx
⇒ \(\frac { dy }{ dx }\) + (sec² x) y = tan x sec² x
which is L.D.E in y of first order and is of the form \(\frac { dy }{ dx }\) + Py = Q
Here, P = sec² x ; Q = tan x sec² x

which gives the required solution.

Question 25.
\(\frac{d y}{d x}=\frac{\left(1+\cos ^2 x\right) \sin ^2 y}{\left(1+\sin ^2 y\right) \cos ^2 x}\)
Solution:

Question 26.
(1 + y + x²y)dx + (x + x³)dy = 0
Solution:
Given (1 + y + x²y) dx + (x + x³) dy = 0

which gives the required solution.

Question 27.
2\(\frac{d y}{d x}=\frac{y}{x}+\frac{y^2}{x^2}\)
Solution:
Given 2\(\frac{d y}{d x}=\frac{y}{x}+\frac{y^2}{x^2}\)

on squaring both sides ; we have
(y – x)² = Axy² ; where c² = A
which gives the required solution.

Question 28.
(1 – x²)\(\frac { dy }{ dx }\) xy = x, given y = 2 when x = 0.
Solution:

Question 29.
\(\frac{d y}{d x}-e^{y+x}=e^{x-y}\)
Solution:

Question 30.
tan x \(\frac { dy }{ dx }\) + 2y = sec x
Solution:
Given tan x \(\frac { dy }{ dx }\) + 2y = sec x
⇒ \(\frac { dy }{ dx }\) + (2 cot x)y = \(\frac { sec x }{ tan x }\) = cosec x
which is linear in y of first order and is of the fonn \(\frac { dy }{ dx }\)+ Py = Q.
Here P = 2 cot x ; Q = cosec x
∴ I.F. = \(e^{\int \mathrm{P} d x}=e^{\int 2 \cot x}=e^{2 \log \sin x}\)
= \(e^{\log \sin ^2 x}=\sin ^2 x\)
and solution is given by
\(y \cdot e^{\int \mathrm{P} d x}=\int \mathrm{Q} \cdot e^{\int \mathrm{P} d x} d x+c\)
y . sin² x = ∫ cosec x. sin² x dx + c
⇒ y sin² x = – cos x + c
which gives the required solution.

Question 31.
x(x² – x²y²)dy + y(y² + x²y²)dx = 0
Solution:

Question 32.
ydx – (x + 2y²)dy = 0
Solution:
Given ydx – (x + 2y²)dy = 0

be the required solution.

Question 33.
tan y dx + sec² y tan x dy = 0
Solution:
Given tan y dx + sec² y tan x dy = 0

which gives the required solution.

Question 34.
(x² + y²)dx – 2xydy = 0, given that y = 0, when x = 1.
Solution:
Given diff. eqn. can be written as

from (2); we have
x² – y² = x be the required solution.

Question 35.
(x cos y)dy = e^x (xlog x + 1)dx
Solution:
Given x cos y dy = e^x (x log x + 1)

gives the required solution.

Question 36.
dy = (5x – 4y) when y = 0, x = 0.
Solution:
Given dy = (5x – 4y) dx

which gives the required particular solution.

Question 37.
cosec³ xdy – cosec y dx = 0.
Solution:
Given cosec³ x dy – cosec y dx = 0

where A = – C
which gives the required solution.

Question 38.
(y + log x) dx – x dy = 0, given that y = 0, when x = 1.
Solution:
Giyen (y + log x) dx – x dy = 0

given that y = 0 when x = 1
from (1); we have
0 = – 1 – log 1 + c ⇒ c = 1
Thus eqn. (1) becomes ; y = – 1 – log x + x
which gives the required particular solution.

Question 39.
\(\frac{d y}{d x}=e^{x+y}+x^2 e^y\)
Solution:

Question 40.
y – x\(\frac { dy }{ dx }\) = x + y\(\frac { dy }{ dx }\), when y = 0 and x = 1.
Solution:
Given y – x\(\frac { dy }{ dx }\) = x + y\(\frac { dy }{ dx }\)

When y = 0 and x = 1 ∴ from (1); we have
log 1 + 2 tan^-1 0 = A ⇒ A = 0
Thus eqn. (1) becomes ;
log (x² + y²) + 2 tan^-1\(\frac { y }{ x }\) = 0
be the required solution.

Question 41.
(xy² + x)dx + (x²y + y)dy = 0.
Solution:
(xy² + x) dx + (x²y + y) dy = 0
⇒ x(y² + 1)dx + y(x² + 1)dy = 0 …(1)
dividing throughout eqn. (1) by (1 + x²) (1 + y²); we get
\(\frac{x d x}{x^2+1}+\frac{y d y}{y^2+1}\); on integrating
\(\int \frac{x d x}{1+x^2}+\int \frac{y d y}{1+y^2}\)
⇒ \(\frac{1}{2} \log \left(1+x^2\right)+\frac{1}{2} \log \left(1+y^2\right)=\frac{1}{2} \log c\)
⇒ log (1 + x²) (1 + y²) = log c
⇒ (1 + x²) (1 + y²) = c
be the required solution.

Question 42.
\(\frac { dy }{ dx }\) – 3y cot x = sin 2x, given y = 2, when x = \(\frac { π }{ 2 }\).
Solution:
Given \(\frac { dy }{ dx }\) – 3y cot x = sin 2x
which is L.D.E in y of first order and is of
the form \(\frac { dy }{ dx }\) + Py = Q
Here P = – 3 cot x ; Q = sin 2x

Given y = 2 when x = \(\frac { π }{ 2 }\)
∴ from (1); we have f
2 = – 2.1 + c.1 ⇒ c = 4
Thus eqn. (1) becomes ;
y = – 2 sin 2x + 4 sin³x
be the required solution.

Question 43.
\(\log \left(\frac{d y}{d x}\right)\) = 2x – 3y.
Solution:

Question 44.
ye^ydx = (y³ + 2xe^y) dy, given there x = 0, y = 1.
Solution:
Given ye^ydx = (y³ + 2xe^y) dy

Given x = 0, y = 1 ∴ from (1); we have
0 = 1 (c – e^-1) ⇒ c = e^-1
Thus eqn. (1) becomes ; x = y² (e^-1 – e^-y)
which gives the required solution.

Question 45.
(x + 1)dy – 2xydx = 0.
Solution:
Given (x + 1) dy – 2xy dx = 0
⇒ \(\frac{1}{y} d y-\frac{2 x d x}{1+x}\) = 0 ; on integrating
\(\int \frac{1}{y} d y-2 \int \frac{x d x}{1+x}\) = 0
⇒ log | y | – 2 [x – log | (1 + x) | ] = c
be the required solution.

Question 46.
(3xy + y²)dx + (x² + xy)dy = 0
Solution:
Given (3xy + y²) dx + (x² + xy) dy = 0

be the required solution.

Question 47.
\(\sin ^{-1}\left(\frac{d y}{d x}\right)\) = x + y
Solution:

Question 48.
\(e^{\frac{x}{y}}\left(1+\frac{x}{y}\right)+\left(1+e^{\frac{x}{y}}\right) \frac{d x}{d y}\) = 0, when x = 0, y = 1.
Solution:

Question 49.
\(x \frac{d y}{d x}+y=3 x^2-2\)
Solution:

Question 50.
x²dy + (xy + y²)dx = 0,
when x = 1 and y = 1.
Solution:
Given x² dy + (xy + y²) dx = 0

given x = 1 when y = 1
∴ from (2); we have
1 = c x 3 ⇒ c = \(\frac { 1 }{ 3 }\)
Thus, eqn. (2) becomes ;
3x²y = y + 2x be the required solution.

Question 51.
The degree of the differential equation
\(\left(\frac{d y}{d x}\right)^4+3 x \frac{d^2 y}{d x^2}\) = 0 is …………..
Solution:
The highest ordered derivative existing in given differen \(\frac{d^2 y}{d x^2}\) and its exponent be 1.
∴ degree of given diff. eqn. be 1.

Question 52.
The degree of the differential equation
\(x\left(\frac{d^2 y}{d x^2}\right)^3+y\left(\frac{d y}{d x}\right)^4+x^3=0\) = 0 is …………..
Solution:
Here highest ordered drerivative existing in given differential eqn by \(\frac { 1 }{ 2 }\) and its power be 3. Thus degree of given diff. eqn be 3.

Question 53.
The degree of the differential equation
\(\frac{d^2 y}{d x^2}+e^{d y / d x}\) = 0.
Solution:
Given differential equation be,
\(\frac{d^2 y}{d x^2}+e^{d y / d x}\) = 0
Since the L.H.S. of differential eqn. cannot be expressed as polynomial in differential coefficients as it contains term like e^dy/dx.
Hence the degree of given differential eqn. is not defined.

Question 54.
The sum of the order and degree of the differential equation
\(\left(\frac{d^2 y}{d x^2}\right)^2+\left(\frac{d y}{d x}\right)^3+x^4\) = 0 is …………..
Solution:
The highest ordered derivative existing in given duff. eqn. be \(\frac{d^2 y}{d x^2}\).
∴ its order be 2. Further, exponent of highest ordered derivative be 2.
∴ degree of given duff. eqn. be 2. Thus required sum = 2 + 2 = 4

Question 55.
The sum of the order and degree of the differential equation
\(\frac{d}{d x}\left\{\left(\frac{d y}{d x}\right)^3\right\}\) = 0 is …………..
Solution:
Given diff. eqn. can be written as ;
\(3\left(\frac{d y}{d x}\right)^2 \frac{d^2 y}{d x^2}\) = 0
Thus, highest ordered derivative existing in given diff. eqn. be \(\frac{d^2 y}{d x^2}\) and its exponent be 1. Thus order of given duff. eqn. be 2 and its degree be I.
∴ required sum = 2 + 1 = 3

Question 56.
The differential equation representing the family of non-horizontal lines y = mx + c in a plane is …………….
Solution:
We know that eqn. of line in plane be ax + by = 1 … (1)
Now eqn. (1) is parallel to x-axis then it is of the form y = constant i.e. a = 0 and b ≠ 0
So. eqn. of all non-horizontal lines in a plane be ax + by = 1, where a ≠ 0 and b ∈ 0
Duff. w.r.t y; we have
\(a \frac{d x}{d y}+b=0 \Rightarrow \mathrm{a} \frac{d^2 x}{d y^2}\) = 0
Since a ≠ 0 ∴\(\frac{d^2 x}{d y^2}\) = 0
Hence its order be 2.

Question 57.
The differential equation obtained by eliminating the arbitrary constant c In the equation representing the family of curves xy = c cos x is …………..
Solution:
Given family of curve be,
xy = c cos x … (1)
duff. eqn. (1) both sides w.r.t. x;
x\(\frac { dy }{ dx }\) + y = – c sin x … (2)
eliminating e from eqn. (1) and (2) ; we have
x\(\frac { dy }{ dx }\) + y = – xy tan x

Question 58.
The solution of the differential equation
\(\frac { dy }{ dx }\) + 2x = e^3x is …………
Solution:
Given diff. eqn. be \(\frac { dy }{ dx }\) + 2x = e^3x
⇒ dy = (e^3x – 2x) dx
On integrating both sides ; we have
y = \(\frac{e^{3 x}}{3}-x^2+\mathrm{C}\)

Question 59.
The equation \(\frac{d y}{d x}+\frac{y}{x \log x}=\frac{1}{x}\) is a ……………. differential equation.
Solution:
Given diff. eqn. be linear diff. eqn. of first order and is of the form \(\frac { dy }{ dx }\) + Py = Q
where P = \(\frac{1}{x \log x}\), Q = \(\frac { 1 }{ x }\)

Question 60.
The integrating factor of \(\frac{d y}{d x}=\frac{x+2 y}{x}\) is …………..
Solution:
Given diff. eqn. can be written as ;
\(\frac{d y}{d x}-\frac{2 y}{x}\) = 1
which is L.D.E of first order and is of the
form \(\frac { dy }{ dx }\) + Py = Q ; where P = – \(\frac { 2 }{ x }\) ; Q = 1
∴ I.F. = \(e^{\int P d x}=e^{\int-\frac{2}{x} d x}\)
= \(e^{-2 \log |x|}=e^{\log |x|^{-2}}=\frac{1}{x^2}\)

Question 61.
The degree of the differential equation
\(\frac{d^2 y}{d x^2}=\left\{1+\left(\frac{d y}{d x}\right)^2\right\}^{3 / 2}\) is
(a) 4
(b) \(\frac { 3 }{ 2 }\)
(c) Not defined
(d) 2
Solution:
Given diff. eqn. be,
\(\frac{d^2 y}{d x^2}=\left[1+\left(\frac{d y}{d x}\right)^2\right]^{3 / 2}\)
On squaring both sides ; we have
\(\left(\frac{d^2 y}{d x^2}\right)^2=\left[1+\left(\frac{d y}{d x}\right)^2\right]^3\)
The highest order derivative present in given diff. eqn. \(\frac{d^2 y}{d x^2}\) and its power is 2.
Thus the order of given diff. eqn. be 2 and degree 2. Clearly it is a non linear differential eqn., as it contains terms like
\(\left(\frac{d^2 y}{d x^2}\right)^2 \text { and }\left(\frac{d y}{d x}\right)^2\).

Question 62.
The curve y = (cos x + y)^1/2 satisfies the differential equation
(a) \((2 y-1) \frac{d^2 y}{d x^2}-2\left(\frac{d y}{d x}\right)^2\) + cos x = 0
(b) \(\frac{d^2 y}{d x^2}-2 y\left(\frac{d y}{d x}\right)^2\) + cos x = 0
(c) \((2 y-1) \frac{d^2 y}{d x^2}-2\left(\frac{d y}{d x}\right)\) + cos x = 0
(d) \((2 y-1) \frac{d^2 y}{d x^2}-\left(\frac{d y}{d x}\right)^2\) + cos x = 0
Solution:
Given eqn. of curve be

Question 63.
If xy = A sin x + B cos x is the solution of the differential equation
\(x \frac{d^2 y}{d x^2}-5 a \frac{d y}{d x}\) + xy = 0,
then the value of a is equal to
(a) \(\frac { 2 }{ 5 }\)
(b) \(\frac { 5 }{ 2 }\)
(c) –\(\frac { 2 }{ 5 }\)
(d) –\(\frac { 5 }{ 2 }\)
Solution:
Given eqn. of soln. be,
xy = A sin x + B cos x … (1)
Diff. eqn. (1) both sides w.r.t. x ;

Question 64.
The solution of differential equation xdy – ydx = 0 represents
(a) a rectangular hyperbola
(b) parabola whose vertex is at origin
(c) straight line passing through origin
(d) a circle whose centre is at origin
Solution:
Given diff. eqn. be, x dy – y dx = 0
⇒ \(\frac{d y}{y}-\frac{d x}{x}\) = 0 ; On integrating
\(\int \frac{d y}{y}-\int \frac{d x}{x}=\log C\)
⇒ log|y| – log|x| = log C
⇒ \(\left|\frac{y}{x}\right|=\mathrm{C}\)
⇒ y = Ax
which represents a straight line passing through origin.

Question 65.
The integrating factor of
x \(\frac { dy }{ dx }\) – y = x⁴ – 3x is
(a) x
(b) log x
(c) \(\frac { 1 }{ x }\)
(d) – x
Solution:
Given diff. eqn. can be written as,

Question 66.
The solution of \(\frac { dy }{ dx }\) + y tanx: = sec x, y (0) = 0 is
(a) y secx = tan x
(b) y tan x = sec x
(c) tan x = y tan x
(d) sec x = tan y.
Solution:
Given diff. eqn. be,
\(\frac { dy }{ dx }\) + y tan x = sec x
which is L.D.E of first order and is of the form
\(\frac { dy }{ dx }\) + Py = Q ; where P = tan x; Q = sec x
and I.F. = \(e^{\int \mathrm{P} d x}=e^{\int \tan x d x}\)
\(e^{-\log |\cos x|}=\sec x\)
and soln. be given by
\(y \cdot e^{\int \mathrm{P} d x}=\int \mathrm{Q} \cdot e^{\int \mathrm{P} d x} d x+\mathrm{C}\)
⇒ ysecx = ∫ sec x. sec x dx + C
⇒ y sec x = tanx + C …(1)
Now y (0) = 0 ∴ from (1); we have
0 x 1 = 0 + C ⇒ C = 0
Thus, eqn. (1) becomes ;
y sec x = tan x

Question 67.
The particular solution of the differential equation x dy + 2y dx = 0, when x = 2, y = 1 is
(а) xy = 4
(b) x²y = 4
(c) xy² = 4
(d) x²y² = 4.
Solution:
Given diff. eqn. be, x dy + 2y dx = 0
after variable separation ; we have
\(\frac { dy }{ dx }\) + 2 \(\frac { dx }{ x }\) = 0
On integrating both sides ; we have
\(\int \frac{d y}{y}+2 \int \frac{d x}{x}=\log C\)
⇒ logy + 2 log x = log C
⇒ x²y = C …(1)
When x = 2, y = 1 from (1); 4 = C
Thus, eqn. (1) becomes ; x²y = 4 be the required particular soln.

Question 68.
The general solution of the differential equation (1 + y²) dx + (1 + x²) dy = 0 is
(a) x – y = c(1 – xy)
(b) x – y = c(1 + xy)
(c) x + y = c(1 – xy)
(d) x + y = c(1 + xy)
Solution:
Given diff. eqn. be written as ;
\(\frac{d x}{1+x^2}+\frac{d y}{1+y^2}\)
On integrating both sides ; we have
\(\int \frac{d x}{1+x^2}+\int \frac{d y}{1+y^2}=\tan ^{-1} c\)
⇒ tan^-1 x + tan^-1 y = tan^-1 c
⇒ \(\tan ^{-1}\left(\frac{x+y}{1-x y}\right)\) = tan^-1 c
⇒ x + y = c(1 – xy)

Question 69.
The solution of the differential equation
\(\frac{d y}{d x}=\frac{x}{y}+\frac{y}{x}\) is
(a) log \(\frac { y }{ x }\) = x² + c
(b) 2 log \(\frac { y }{ x }\) = x² + c
(c) (\(\frac { y }{ x }\))² = log x + c
(d) (\(\frac { y }{ x }\))² = 2 log x + c
Solution:
Given differential eqn. can be written as ;

Question 70.
The differential equation y \(\frac { dy }{ dx }\) + x = a (a is a constant) represents
(a) a set of circles having centre on they- axis
(b) a set of circles having centre on the x-axis
(c) a set of ellipses
(d) None of these
Solution:
Given diff. eqn. be, y \(\frac { dy }{ dx }\) + x = a
⇒ y dy = (a – x) dx
On integrating both sides ; we have
\(\frac{y^2}{2}=\frac{(a-x)^2}{-2}+\frac{C}{2}\)
⇒ (x – a)² + y² = C
which represents a family of circles having centre on x-axis.

Question 71.
If p and q are the order and degree respectively of the differential equation
y\(\frac { dy }{ dx }\) + x³\(\left(\frac{d^2 y}{d x^2}\right)\)³ + xy = cos x, then
(a) p < q (b) p = q (c) p > q
(d) None of these
Solution:
Here the highest ordered derivative existing in given diff. eqn. be
\(\frac{d^2 y}{d x^2}\) and its power be 3.
Thus order of given diff. eqn. be 2 i.e. p = 2
and degree of given diff. eqn. be 3 i.e. q = 3
∴ q > p

Question 72.
The degree and order of the differential equation \(\left[1+\left(\frac{d y}{d x}\right)^3\right]^{7 / 3}=\left(7 \frac{d^2 y}{d x^2}\right)\) respectively are
(a) 3 and 7
(b) 3 and 2
(c) 7 and 3
(d) 2 and 3
Solution:
Given diff. eqn. can be written as ;
\(\left[1+\left(\frac{d y}{d x}\right)^3\right]^7=\left(7 \frac{d^2 y}{d x^2}\right)^3\)
The highest ordered derivative existing is given diff. eqn. be \(\frac{d^2 y}{d x^2}\) and its exponents be 3.
Therefore order of given diff. eqn. be 2 and its degree be 3.

Question 73.
The order and degree of the differential equation y = \(x \frac{d y}{d x}+\frac{2}{\frac{d y}{d x}}\) is
(a) 1, 2
(b) 1, 3
(c) 2, 1
(d) 1, 1
Solution:
Given diff. eqn. can be written as ;
\(y \frac{d y}{d x}=x\left(\frac{d y}{d x}\right)^2\) + 2
Here the highest ordered derivative existing in given diff. eqn. be \(\frac { dy }{ dx }\) and its exponent be 2.
Thus order of given diff. eqn. be 1 and its degree be 2.

Question 74.
The degree and order of the differential equation y = \(x\left(\frac{d y}{d x}\right)^2+\left(\frac{d x}{d y}\right)^2\) respectively
(a) 1, 1
(b) 2, 1
(c) 4, 1
(d) 1, 4
Solution:
Given diff. eqn. be, y = \(x\left(\frac{d y}{d x}\right)^2+\left(\frac{d x}{d y}\right)^2\)
⇒ y = x \(\left(\frac{d y}{d x}\right)^2+\frac{1}{\left(\frac{d y}{d x}\right)^2}\)
⇒ \(y\left(\frac{d y}{d x}\right)^2=x\left(\frac{d y}{d x}\right)^4+1\)
which is diff. eqn. in x and y. The highest ordered derivative existing in diff. eqn. be \(\frac { dy }{ dx }\) and its power be 4.
Thus, degree of given differential eqn. be 4 and the order of differential equation be 1.

Question 75.
The degree of the differential equation
\(\frac{d^2 y}{d x^2}=\frac{5 y+\frac{d y}{d x}}{\sqrt{\frac{d^2 y}{d x^2}}}\)
(a) 2
(b) 3
(c) 4
(d) 5/2
Solution:
Given diff. eqn. can be written as,
\(\left(\frac{d^2 y}{d x^2}\right)^{3 / 2}=5 y+\frac{d y}{d x}\)
⇒ \(\left(\frac{d^2 y}{d x^2}\right)^3=\left(5 y+\frac{d y}{d x}\right)^2\)
Here the highest ordered derivative existing be in given diff. eqn. be \(\frac{d^2 y}{d x^2}\) and its exponent be 3.
Thus degree of given diff. eqn. be 3.

Question 76.
The order and degree of the differential equation \(\frac{d^2 y}{d x^2}+y+\left(\frac{d y}{d x}+\frac{d^3 y}{d x^3}\right)^{5 / 2}\) = 0 respectively are
(a) 3, 2
(b) 2, 3
(c) 3, 1
(d) 3, 5
Solution:
Given diff. eqn. can be written as ;
\(\left(\frac{d^2 y}{d x^2}+y\right)^2=\left(\frac{d y}{d x}+\frac{d^3 y}{d x^3}\right)^5\)
The highest ordered derivative existing in given diff. eqn. be \(\frac{d^3 y}{d x^3}\) and its exponent be 5. Thus order of given diff. eqn. be 3 and its degree be 5.

Question 77.
A solution of the differential equation
\(\left(\frac{d y}{d x}\right)^2-x \frac{d y}{d x}\) + y = 0 is
(a) y = 2x
(b) y = – 2x
(c) y = 2x – 4
(d) y = 2x + 4
Solution:

Question 78.
The integrating factor of
cos x \(\frac { dy }{ dx }\) + y sin x = 1 is
(a) cos x
(b) tan x
(c) sec x
(d) sin x
Solution:
Given diff. eqn. can be written as ;
\(\frac { dy }{ dx }\) + y tan x = sec x
which is L.D.E of first order and is of the form
\(\frac { dy }{ dx }\) + Py = Q ; where P = tan x ; Q = sec x
∴ I.F. = \(e^{\int P d x}=e^{\int \tan x d x}=e^{-\log |\cos x|}\)
= \(e^{\log |\cos x|^{-1}}=\frac{1}{\cos x}=\sec x\)

Question 79.
Find the degree of the differential equation 1 + (\(\frac { dy }{ dx }\))² = x
Solution:
The highest ordered derivative existing in given diff. eqn. be \(\frac { dy }{ dx }\) and its exponent be 2.
Thus, degree of given diff. eqn. be 2.

Question 80.
Find the order and degree of the differential equation
\(x^2 \frac{d^2 y}{d x^2}=\left\{1+\left(\frac{d y}{d x}\right)^2\right\}^4\)?
Solution:
Given differential eqn. be,
\(x^2 \frac{d^2 y}{d x^2}=\left\{1+\left(\frac{d y}{d x}\right)^2\right\}^4\)
which represents a polynomial in derivatives.
The highest ordered derivative existing in diff. eqn. be \(\frac { d²y }{ dx² }\) and its power be 1. Thus
degree of given diff. eqn. be 1 and order 2. Since given diff. eqn. contains term like (\(\frac { dy }{ dx }\))²
∴ given differential eqn. is non-linear.

Question 81.
Find the order and degree of the differential equation \(x^3\left(\frac{d^2 y}{d x^2}\right)^2+x\left(\frac{d y}{d x}\right)^4\) = 0
Solution:
Here the highest ordered derivative existing in differential equation be \(\frac { d²y }{ dx² }\) and its power is 2. ∴ Order of the differential equation be 2. Thus degree of given differential equation be also 2.

Question 82.
Form the differential equation representing the family of curves
(i) y = \(\frac { A }{ x }\) + 5
(ii) y = A sin x
(iii) xy = c cos x
(iv) y = mx
(v) \(\frac { A }{ r }\) + b, where A and B are arbitrary constants.
Solution:
(i) Given eqn. of family of curves be

(iii) Given family of curve be,
xy = c cos x …(1)
diff. eqn. (1) both sides w.r.t. x ;
x \(\frac { dy }{ dx }\) + y = – c sinx …(2)
eliminating c from eqn. (1) and (2) ; we have
x\(\frac { dy }{ dx }\) + y = – xy tan x

(iv) Given eqn. of family of curves be, dy
y = mx
∴ \(\frac { dy }{ dx }\) = m
⇒ \(\frac { dy }{ dx }\) = \(\frac { y }{ x }\) which is the required diff. eqn.

(v) Given eqn. of family of curves be,

which is the required diff. eqn.

Question 83.
Write the general solution of differential equation
(i) \(\frac{d y}{d x}=e^{x+y}\)
(ii) \(\frac{d y}{d x}=x^3 e^{-2 y}\)
Solution:
(i) Given diff. eqn be, \(\frac { dy }{ dx }\) = e^x+y
⇒ \(\frac { dy }{ dx }\) = e^x . e^y ⇒ e^-y dy = e^x dx
on integrating; we have
– e^-y = e^x + c
⇒ e^-y + e^x = A be the required solution.

(ii) Given diff. eqn. be,
\(\frac { dy }{ dx }\) = x³e^-2y ⇒ e^2y dy = x³ dx
On integrating both sides ; we have
\(\frac{e^{2 y}}{2}=\frac{x^4}{4}+\mathrm{C}\)
which is the required solution.

Question 84.
Show that the function y = ax + 2a² is a solution of differential equation
\(2\left(\frac{d y}{d x}\right)^2+x\left(\frac{d y}{d x}\right)\) – y = 0
Solution:
Eqn. of given function be, y = ax + 2a² … (1)
∴ \(\frac { dy }{ dx }\) = a
∴ \(2\left(\frac{d y}{d x}\right)^2+x \frac{d y}{d x}\) – y
= 2a² + ax – ax – 2a² = 0
Thus, eqn. (1) be the soln. of given diff. eqn, \(2\left(\frac{d y}{d x}\right)^2+x \frac{d y}{d x}-y\) = 0

Question 85.
What form of the differential equation is the equation \(\frac{d y}{d x}+\frac{y}{x \log x}=\frac{1}{x}\)
Solution:
Given diff. eqn. be linear diff. eqn. of first order and is of the form \(\frac { dy }{ dx }\) + Py = Q
where P = \(\frac{1}{x \log x}\) ; Q = \(\frac { 1 }{ x }\)

Question 86.
Write the integrating factor of the differential equation :
(i) x \(\frac { dy }{ dx }\) + 2y = x²
(ii) \(\left(\tan ^{-1} y-x\right) d y=\left(1+y^2\right) d x\)
(iii) \(\left(\frac{e^{-2 \sqrt{x}}}{\sqrt{x}}-\frac{y}{\sqrt{x}}\right) \frac{d x}{d y}\)
Answer:
(i) Given differential equation can be written as \(\frac { dy }{ dx }\) + \(\frac { 2 }{ x }\)y = x
which is first order L.D.E. in y and is of the form \(\frac { dy }{ dx }\) + Py = Q ; where P = \(\frac { 2 }{ x }\) and Q = x