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

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

Question 1.
Verify that the function y = e^ax (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 = e^ax (c₁ cos bx + c₂ sin bx) …………….(1)
Diff. (1) w.r.t. x we have
\(\frac{d y}{d x}\) = ae^ax (c₁ cos bx + c₂ sin bx) + e^ax (- c₁ b sin bx + c₂ b cos bx)
⇒ \(\frac{d y}{d x}\) = ay + be^ax (- 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 {e^ax {- c₁ b cos bx – c₂ b sin bx) + (- c₁ sin bx + c₂ cos bx) ae^ax}
⇒ \(\frac{d^2 y}{d x^2}\) = 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 = e^ax (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 log_e 2
Hence, the amount will be double after 20 log_e 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}
= e^y
and solution is given by
x . e^{∫ P dy} = ∫ Q . e^{∫ P dy} dy + C
⇒ x . e^y = ∫ e^{– y} sec² y e^y dy + C
⇒ x . e^y = 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}
= \(e^{\int \frac{2}{y} d y}\)
= e^{2 log |y|}
= e^{log 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 log_e 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 = e^x2
[∵ 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

⇒ xe^{sin-1 y} = ∫ t e^t dt + C
= (t – 1) e^t + C
⇒ xe^{sin-1 y} = (sin^-1 y – 1) e^{sin-1 y} + C ……………..(1)
When x = 0, y = 0
∴ from (1) ;
0 = (0 – 1) + C
⇒ C = 1
∴ from (1) ; we get
xe^{sin-1 y} = (sin^-1 y – 1) e^{sin-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.