Utilizing ML Aggarwal Class 12 ISC Solutions Chapter 9 Differential Equations Chapter Test as a study aid can enhance exam preparation.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 9 Differential Equations Chapter Test

Question 1.
Verify that the function y = eax (c1 cos bx + c2 sin bx), where c1, c2 are arbitrary constants is a solution of the differential equation $$\frac{d^2 y}{d x^2}$$ – 2a $$\frac{d y}{d x}$$ + (a2 + b2) y = 0. (NCERT)
Solution:
Given,
y = eax (c1 cos bx + c2 sin bx) …………….(1)
Diff. (1) w.r.t. x we have
$$\frac{d y}{d x}$$ = aeax (c1 cos bx + c2 sin bx) + eax (- c1 b sin bx + c2 b cos bx)
⇒ $$\frac{d y}{d x}$$ = ay + beax (- c1 sin bx + c2 cos bx)
Diff. eqn. (2) w.r.t. x ; we have
$$\frac{d^2 y}{d x^2}$$ = a $$\frac{d y}{d x}$$ + b {eax {- c1 b cos bx – c2 b sin bx) + (- c1 sin bx + c2 cos bx) aeax}
⇒ $$\frac{d^2 y}{d x^2}$$ = a $$\frac{d y}{d x}$$ – b2y + a $$\frac{d y}{d x}$$ – a2y
[using eqn. (1) and (2)]
⇒ $$\frac{d^2 y}{d x^2}$$ – 2a $$\frac{d y}{d x}$$ + (a2 + b2) y = 0
which is the given diff. eqn.
Hence, y = eax (C1 cos bx + C2 sin bx) be the solution of given diff. eqn.

Question 1 (old).
Show that y = $$\frac{c-x}{1+c x}$$, is arbitrary constant, is a solution of the differential equation (1 + x2) $$\frac{d y}{d x}$$ + (1 + y2) = 0.
Solution:
Given y = $$\frac{C-x}{1+C x}$$
⇒ y (1 + Cx) = C – x
⇒ C (xy – 1) = – x – y
⇒ C = $$\frac{x+y}{1-x y}$$ ……………..(1)
Diff. (1) w.r.t. x; we have
$$\frac{(1-x y)\left(1+\frac{d y}{d x}\right)+(x+y)\left(x \frac{d y}{d x}+y\right)}{(1-x y)^2}$$ = 0
⇒ (1 + x2) $$\frac{d y}{d x}$$ + 1 + y2 = 0, which is given diff. eqn.
Thus, y = $$\frac{c-x}{1+c x}$$ be the required solution of given diff. eqn.

Question 2.
Form the differential equation of the family of curves represented by $$\sqrt{1-x^2}+\sqrt{1-y^2}$$ = a (x – y), a being parameter.
Solution:
Given family of curves represented by
$$\sqrt{1-x^2}+\sqrt{1-y^2}$$ = a (x – y)
put x = sin θ
⇒ θ = sin-1 x
and y = sin Φ
Φ = sin-1 y
∴ From (1) ; we have
$$\sqrt{1-\sin ^2 \theta}+\sqrt{1-\sin ^2 \theta}$$ = a (sin θ – sin Φ)
⇒ cos θ + cos Φ = a (sin Φ – sin Φ)
⇒ $$2 \cos \frac{\theta+\phi}{2} \cos \frac{\theta-\phi}{2}=2 a \cos \frac{\theta-\phi}{2} \sin \frac{\theta+\phi}{2}$$
⇒ cot $$\frac{\theta+\phi}{2}$$ = a
⇒ θ + Φ = 2 cot-1 a
⇒ sin-1 x + sin-1 y = 2 cot-1 a
Diff. both sides w.r.t. x; we have
$$\frac{1}{\sqrt{1-x^2}}+\frac{1}{\sqrt{1-y^2}} \frac{d y}{d x}$$ = 0
which gives the required differential equation.

Question 3.
Obtain the differential equation of the family of straight lines which are at a fixed distance p from the origin.
Solution:
Now, equation of family of straight lines which are at a fixed distance p from the origin is given by
x cos α + y sin α = p ………………..(1)
where α be the parameter
On differentiating eqn. (1) w.r.t. x; we have
cos α + sin α $$\frac{d y}{d x}$$ = 0 ……………..(2)
eqn. (1) – x × eqn. (2) ; we have
sin α (y – x $$\frac{d y}{d x}$$) = p ……………….(3)
eqn. (1) × $$\frac{d y}{d x}$$ – eqn. (2) × y ; we have
(x $$\frac{d y}{d x}$$ – y) cos α = p $$\frac{d y}{d x}$$ ………………(4)
On squaring and adding eqn. (3) and eqn. (4) ; we have
(xy1 – y)2 (cos2 α + sin2 α) = p2 (1 + ($$\frac{d y}{d x}$$)2]
⇒ (y – xy1)2 = p2 (1 + y12)
which gives the required differential eqn.

Question 4.
(tan2 x + 2 tan x + 5) = 2 (1 + tan x) sec2 x.
Solution:
Given, (tan2 x + 2 tan x + 5) = 2 (1 + tan x) sec2 x
⇒ dy = $$\frac{2(1+\tan x) \sec ^2 x}{\tan ^2 x+2 \tan x+5}$$ dx
[after variable separation]
On integrating ; we have
∫ dy = ∫ $$\frac{2(1+\tan x) \sec ^2 x d x}{\tan ^2 x+2 \tan x+5}$$ + C ……………….(1)
put tan2 x + 2 tan x + 5 = t
⇒ (2 tan x sec2 x + 2 sec2 x) dx = dt
⇒ 2 sec2 x (1 + tan x) dx = dt

Question 5.
(i) x (y2 – a) dx + y (1 + x) dy = 0
(ii) cos x (1 + cos y) dy – sin y (1 + sin x) dx = 0
Solution:
(i) Given, x (y2 – a) dx + y (1 + x) dy = 0
⇒ $$\frac{x d x}{1+x}+\frac{y d y}{y^2-a}$$ = 0
[after variable separation]
On integrating ; we have
$$\int \frac{1+x-1}{1+x} d x+\int \frac{y d y}{y^2-a}$$ = C
⇒ $$\int\left[1-\frac{1}{1+x}\right] d x+\frac{1}{2} \int \frac{2 y d y}{y^2-a}$$ = C
⇒ x – log |1 + x| + $$\frac{1}{2}$$ log |y2 – a| = C
which gives the required solution.

(ii) Given, cos x (1 + cos y) dy – sin y (1 + sin x) dx = 0

Question 5 (old).
Find the differential equation of all parabolas having their axis of symmetry as the x-axis.
Solution:
The equation of all parabolas having axis of symmetry as the x-axis be given by
(y – 0)2 = 4a (x – h) ……………..(1)
where a, h are arbitrary constants
Diff. eqn. (1) both sides w.r.t. x; we have
2yy1 = 4a
⇒ yy1 = 2a ……………..(2)
again differentiating eqn. (2) w.r.t. x, we get
yy2 + y1 = 0
which gives the required differential equation.

Question 6.
(i) $$\frac{d y}{d x}$$ = x5 tan-1 (x3)
(ii) a (x $$\frac{d y}{d x}$$ + 2y) = xy $$\frac{d y}{d x}$$
Solution:
(i) Given, $$\frac{d y}{d x}$$ = x5 tan-1 (x3)
⇒ dy = x5 tan-1 (x3) dx
On integrating ; we have
∫ dy = ∫ x5 tan-1 (x3) dx
put x3 = t
⇒ 3x2 dx = dt
y = ∫ t tan-1 t $$\frac{d t}{3}$$ + C

(ii) Given a (x $$\frac{d y}{d x}$$ + 2y) = xy $$\frac{d y}{d x}$$
⇒ x (a – y) $$\frac{d y}{d x}$$ = – 2 ay
$$\frac{(a-y)}{a y}=-2 \frac{d x}{x}$$
On integrating ; we have
$$\frac{1}{a} \int\left[\frac{a}{y}-1\right] d y=-2 \int \frac{d x}{x}+\mathrm{C}$$
⇒ $$\frac{1}{a}$$ [a log |y| – y] = – 2 log |x| + C
⇒ log |x2y| = C + $$\frac{y}{a}$$
⇒ a log |yx2| = y + A
which gives the required general solution.

Question 7.
cos x cos y $$\frac{d y}{d x}$$ + sin x sin y = 0.
Solution:
Given, cos x cos y $$\frac{d y}{d x}$$ + sin x sin y = 0
$$\frac{\cos y}{\sin y} d y+\frac{\sin x}{\cos x} d x$$ = 0
On integrating ; we have
$$\int \frac{\cos y}{\sin y} d y+\int \frac{\sin x}{\cos x} d x$$ = log C
⇒ log |sin y| – log |cos x| = log C
⇒ |sin x| = C |cos x|
⇒ sin y = A cos x ; where ± C = A
which gives the required general solution.

Question 8.
Find the particular solution of (1 – x) dy – (1 + y) dy = 0, it being given that y = 4 when x = 2.
Solution:
Given, (1 – x) dy – (1 + y) dy = 0
⇒ $$\frac{d y}{1+y}-\frac{d x}{1-x}$$ = 0
On integrating; we have
$$\int \frac{d y}{1+y}-\int \frac{d x}{1-x}$$ = log C
⇒ log (1 + y) + log (1 – x) = log C
⇒ (1 + y) (1 – x) = C
Thus, eqn. (1) gives the general solution of given differential eqn.
Given y = 4 when x = 2
∴ from (1) ; we have
5(- 1) = C
⇒ C = – 5
Thus from eqn. (1) ; we have
(1 + y) (1 – x) + 5 = 0
which gives the required particular solution of given diff. eqn.

Question 9.
Find the particular solution of the differential equation $$e^{\frac{d y}{d x}}$$ = x + 1, given that y = 3 when x = 0.
Solution:
Given $$e^{\frac{d y}{d x}}$$ = log (x + 1)
$$\frac{d y}{d x}$$ = log (x + 1) dx ;
on integrating ∫ dy = ∫ log (x + 1) . 1 dx
⇒ y = log (x + 1) . x – $$\int \frac{1}{x+1}$$ x dx
⇒ y = x log |(x + 1)| – ∫ $$\left[1-\frac{1}{x+1}\right]$$ dx + c
⇒ y = x log (x + 1) – x + log (x + 1) + C
⇒ y = (x + 1) log |(x + 1)| – x + c ………….(1)
Given y = 3 when x = 0
∴ from (1) ; we have
3 = log 1 – 0 + c
⇒ c = 3
From (1) ; we have
∴ y = (x + 1) log |x + 1| – x + 3
which is the required solution.

Question 10.
Find the equation of the curve passing through the point (0, $$\frac{\pi}{4}$$) whose differential equation is sin x cos y dx + cos x sin y dy = 0. (NCERT)
Solution:
Given differential equation be,
sin x cos y dx + cos x sin y dy = 0
⇒ $$\frac{\sin x}{\cos x}$$ dx + $$\frac{\sin y}{\cos y}$$ dy = 0
On integrating ; we have
$$\int \frac{\sin x d x}{\cos x}+\int \frac{\sin y d y}{\cos y}$$ = – log C
⇒ – log |cos x| – log |cos y| = – log C
⇒ log |cos x cos y| = log C
⇒ cos x cos y = ± C = A ……………(1)
Since the curve (1) passes through (0, $$\frac{\pi}{4}$$).
i.e. When x = 0, y = $$\frac{\pi}{4}$$
∴ from (1) ;
$$1 \times \frac{1}{\sqrt{2}}$$ = A
⇒ A = $$\frac{1}{\sqrt{2}}$$
Thus eqn. (1) gives ; cos x cosy = $$\frac{1}{\sqrt{2}}$$ which gives the required eqn. of curve.

Question 11.
In a bank principal increases at the rate of 5 % per year. In how many years Rs. 1000 double itself. (NCERT)
Solution:
Let P be the principal at any instant t
Then $$\frac{d \mathrm{P}}{d t}=\frac{5}{100} \mathrm{P}$$
⇒ $$\frac{\mathrm{dP}}{\mathrm{P}}=\frac{d t}{20}$$
on integrating; we have
log P = $$\frac{1}{20}$$ t + c …………….(1)
initially at t = 0 ; P = 1000
∴ from (1) ; we have
log 1000 = $$\frac{1 \times 0}{20}$$ + c
⇒ c = log 1000
∴ from eqn (1) ; we have
log P = $$\frac{t}{20}$$ + log (1000)
log $$\frac{\mathrm{P}}{1000}=\frac{t}{20}$$ …………………..(2)
Let t = t1 when P = 2000
∴ from eqn. (2); we have,
log $$\left(\frac{2000}{1000}\right)=\frac{t_1}{20}$$
⇒ t1 = 20 loge 2
Hence, the amount will be double after 20 loge 2 years.

Question 12.
(i) x (x – y) dy + y2 dx = 0
(ii) x3 $$\frac{d y}{d x}$$ = y3 + y2 $$\sqrt{y^2-x^2}$$, x > 0
Solution:
(i) Given, x (x – y) dy + y2 dx = 0

(ii) Given, x3 $$\frac{d y}{d x}$$ = y3 + y2 $$\sqrt{y^2-x^2}$$

Question 13.
(i) $$\left(\frac{e^{-2 \sqrt{x}}}{\sqrt{x}}-\frac{y}{\sqrt{x}}\right) \frac{d x}{d y}$$ = 1, x > 0
(ii) e-y sec2 y dy = dx + x dy
Solution:
(i) Given, $$\left(\frac{e^{-2 \sqrt{x}}}{\sqrt{x}}-\frac{y}{\sqrt{x}}\right) \frac{d x}{d y}$$ = 1

(ii) Given, e-y sec2 y dy = dx + x dy
⇒ (e-y sec2 y – x) dy = dx
⇒ $$\frac{d x}{d y}$$ + y = e-y sec2 y
which is L.D.E. in x and is of the form
$$\frac{d x}{d y}$$ + Px = Q
where P = 1 and Q = e-y sec2 y
∴ I.F. = e∫ P dy
= e∫ dy
= ey
and solution is given by
x . e∫ P dy = ∫ Q . e∫ P dy dy + C
⇒ x . ey = ∫ e– y sec2 y ey dy + C
⇒ x . ey = tan y + C
⇒ x = e– y (tan y + C)

Question 14.
(i) (2x – 10y3) dy + y dx = 0, y ≠ 0
(ii) (x + tan y) dy = sin 2y dx
Solution:
(i) Given, (2x – 10y3) dy + y dx = 0
⇒ $$\frac{d x}{d y}$$ + $$\frac{2 x}{y}$$ – 10 y2 = 0
⇒ $$\frac{d x}{d y}$$ + $$\frac{2 x}{y}$$ = 10 y2
which is linear in x and its of the form
⇒ $$\frac{d x}{d y}$$ + Px = Q ;
where P = $$\frac{2}{y}$$
Q = 10 y2
∴ I.F. = e∫ P dy
= $$e^{\int \frac{2}{y} d y}$$
= e2 log |y|
= elog y2
= y2
and solution is given by
x . e∫ P dy = ∫ Q . e∫ P dy dy + C
⇒ x . y2 = ∫ 10 y2 × y2 dy + C
⇒ xy2 = 10 + C
⇒ xy2 = 2y5 + C, which gives the required solution.

(ii) Given (x + tan y) dy = sin 2y dx

Question 14 (old).
In a bank principal increases at the rate of r % per year. Find the value of r if Rs. 100 double itself in 10 years (given loge 2 = 0.6931). (NCERT)
Solution:
Let P be the pnncipal at any instant t
Then $$\frac{d \mathrm{P}}{d t}=\frac{r}{100} \mathrm{P}$$
⇒ $$\frac{d \mathrm{P}}{P}=\frac{r}{100}$$ dt ;
on integrating ; we get
⇒ log P = $$\frac{r}{100}$$ t + c ……………..(1)
initially at t = 0, P = P0
∴ from (1) ; we have
log P0 = c
∴ from (1) ; we have
log P = $$\frac{r t}{100}$$ + log P0
⇒ $$\left(\frac{p}{P_0}\right)=\frac{r t}{100}$$ …………….(2)
Given P0 = 100 ; P = 200 at t = 10
∴ from (2) ; we have
⇒ $$\left(\frac{200}{100}\right)=\frac{r \times 10}{100}$$
⇒ log 2 = $$\frac{r}{10}$$
⇒ r = 10 × log 2 = 10 × 0.6931 = 6.931
Thus the required rate percent be 6.931% per annum.

Question 15.
Find the particular solutions of the following differential equations :
(i) $$\frac{d y}{d x}$$ = 2xy, given that y = 1 when x = 0.
(ii) 2x2 $$\frac{d y}{d x}$$ – 2xy + y2 = 0, given that y = e when x = e.
(iii) (xy – y2) dx – x2 dy = 0, given that y = 1 when x = 1.
(iv) (x2 + 1) $$\frac{d y}{d x}$$ – 2xy = (x4 + 2x2 + 1) cos x, given that y = 0 When x = 0.
(v) $$\sqrt{1-y^2}$$ dx = (sin-1 y – x) dy, given that y = 0 when x = 0.
(vi) $$x e^{\frac{y}{x}}-y \sin \left(\frac{y}{x}\right)+x \sin \left(\frac{y}{x}\right) \frac{d y}{d x}$$ = 0, given that y = 0 when x = 1.
Solution:
(i) Given, $$\frac{d y}{d x}$$ = 2xy
⇒ $$\frac{1}{y}$$ dy = 2x dx
On integrating ; we have
log |y| = x2 + C ………………..(1)
given, y = 1 when x = 0
∴ from (1) ;
0 = 0 + C
⇒ C = 0
Thus eqn. (1) gives ;
log |y| = x2
⇒ y = ex2
[∵ y > 0]
which gives the required particular solution.

(ii) Given, 2x2 $$\frac{d y}{d x}$$ – 2xy + y2 = 0

⇒ – $$\frac{2 x}{y}$$ + log |x| = C …………….(1)
When x = e, y = e
∴ from (1) ;
– $$\frac{2 e}{e}$$ + log |e| = C
⇒ – 2 + 1 = C
⇒ C = – 1
Thus, eqn. (1) becomes ;
– $$\frac{2 x}{y}$$ + log |x| = – 1
⇒ 1 + log |x| = $$\frac{2 x}{y}$$
⇒ y = $$\frac{2 x}{1+\log |x|}$$ be the required particular solution.

(iii) Given, (xy – y2) dx – x2 dy = 0
⇒ $$\frac{d y}{d x}=\frac{x y-y^2}{x^2}$$ …………..(1)
Also, $$\frac{d y}{d x}=\frac{y}{x}-\left(\frac{y}{x}\right)^2=\phi\left(\frac{y}{x}\right)$$
Thus, eqn. (1) be a homogeneous differential eqn.
put y = vx
⇒ $$\frac{d y}{d x}$$ = v + x $$\frac{d v}{d x}$$
∴ from (1) ;

(iv) Given diff. eqn. can be written as ;

(v) Given $$\sqrt{1-y^2}$$ dx = (sin-1 y – x) dy
⇒ $$\frac{d x}{d y}+\frac{x}{\sqrt{1-y^2}}=\frac{\sin ^{-1} y}{\sqrt{1-y^2}}$$
which is L.D.E. in x and is of the form

⇒ xesin-1 y = ∫ t et dt + C
= (t – 1) et + C
⇒ xesin-1 y = (sin-1 y – 1) esin-1 y + C ……………..(1)
When x = 0, y = 0
∴ from (1) ;
0 = (0 – 1) + C
⇒ C = 1
∴ from (1) ; we get
xesin-1 y = (sin-1 y – 1) esin-1 y + 1
⇒ x = sin-1 y – 1 + e– sin-1 y
which gives the required particular solution.

(vi) Given diff. eqn. can be written as

∴ I = – sin v e-v + [- cos v e-v – ∫ sin v e-v dv]
⇒ 2I = – (sin v + cos v) e-v
∴ from (2) ; we get
$$\frac{-(\sin v+\cos v) e^{-v}}{2}$$ = – log |x| + C
⇒ e-y/x (sin $$\frac{y}{x}$$ + cos $$\frac{y}{x}$$) = 2 log |x| + A …………….(3)
where A = – 2C
When x = 1, y = 0
∴ from (3) ; we have
1 (0 + 1) = 2 × 0 + A
⇒ A = 1
Thus eqn. (4) becomes ;
e-y/x (sin $$\frac{y}{x}$$ cos $$\frac{y}{x}$$) = 2 log |x| + 1
which gives the required particular solution of given diff. equation.