Practicing the AP Inter 2nd Year Maths Study Material and AP Inter 2nd Year Maths Chapter 9 Differential Equations Solutions Exercise 9f will help students to clear their doubts quickly.
Inter 2nd Year Maths Differential Equations Solutions Exercise 9f
Differential Equations Exercise 9f Solutions
I. For each of the differential equations given below, indicate its order and degree (if defined).
Question 1.
\(\frac{d^2 y}{d x^2}+5 x\left(\frac{d y}{d x}\right)^2\) – 6y = log x
Solution:
The given differential equation is
\(\frac{d^2 y}{d x^2}+5 x\left(\frac{d y}{d x}\right)^2\) – 6y = log x
⇒ \(\frac{d^2 y}{d x^2}+5 x\left(\frac{d y}{d x}\right)^2-6 y-\log x=0\)
The highest order derivative is 2.
∴ Order = 2
The index of the highest-order derivative is one.
∴ Degree = 1
Question 2.
\(\left(\frac{d y}{d x}\right)^3-4\left(\frac{d y}{d x}\right)^2\) + 7y = sin x
Solution:
The given differential equation is
\(\left(\frac{d y}{d x}\right)^3-4\left(\frac{d y}{d x}\right)^2\) + 7y = sin x
⇒ \(\left(\frac{d y}{d x}\right)^3-4\left(\frac{d y}{d x}\right)^2\) + 7y – sin x = 0
The highest order derivative is 1.
∴ Order = 1
The index of the highest power is 3.
∴ Degree = 3.
Question 3.
\(\frac{d^4 y}{d x^4}-\sin \left(\frac{d^3 y}{d x^3}\right)=0\)
Solution:
The given differential equation is
\(\frac{d^4 y}{d x^4}-\sin \left(\frac{d^3 y}{d x^3}\right)=0\)
The highest-order derivative is 4.
∴ Order = 4
Since it is not a polynomial equation, the degree is not defined.
Question 4.
For each of the following, verify that the given function (implicit or explicit) is a solution of the corresponding differential equation.
Solution:
Given that \(x \frac{d^2 y}{d x^2}+2 \frac{d y}{d x}-x y+x^2-2=0\) …..(i)
xy = aex + be-x + x2 ……(ii)
Differentiating (ii) w.r.t. x, we get
\(x \frac{d y}{d x}+y=a e^x-b e^{-x}+2 x\)
Again differentiating w.r.t. x, we get
\(x \frac{d^2 y}{d x^2}+\frac{d y}{d x}+\frac{d y}{d x}=a e^x+b e^{-x}+2\)
\(x \frac{d^2 y}{d x^2}+2 \frac{d y}{d x}=x y-x^2+2\) [∵ from (ii)]
\(x \frac{d^2 y}{d x^2}+2 \frac{d y}{d x}-x y+x^2-2=0\)
Hence, the given function is a solution.
Question 5.
y = ex (a cos x + b sin x); \(\frac{d^2 y}{d x^2}-2 \frac{d y}{d x}+2 y=0\)
Solution:
Given that y = ex (a cos x + b sin x)
On dividing by ex on both sides, we get
e-x y = a cos x + b sin x ……(1)
On differentiating both sides w.r.t. x, we get
\(e^{-x} \frac{d y}{d x}-y e^{-x}\) = -a sin x + b cos x
Again, differentiating on both sides w.r.t. x, we get
\(e^{-x} \frac{d^2 y}{d x^2}-\frac{d y}{d x} e^{-x}+y e^{-x}-e^{-x} \frac{d y}{d x}\) = -(a cos x + b sin x)
⇒ \(e^{-x} \frac{d^2 y}{d x^2}-2 e^{-x} \frac{d y}{d x}+y e^{-x}=-y e^{-x}\) (∵ from 1)
⇒ \(e^{-x} \frac{d^2 y}{d x^2}-2 e^{-x} \frac{d y}{d x}+2 y e^{-x}=0\)
⇒ \(e^{-x}\left\{\frac{d^2 y}{d x^2}-2 \frac{d y}{d x}+2 y\right\}=0\)
⇒ \(\frac{d^2 y}{d x^2}-2 \frac{d y}{d x}+2 y=0\)
∴ The given function is a solution of the corresponding differential equation.
Question 6.
y = x sin 3x; \(\frac{d^2 y}{d x^2}\) + 9y – 6 cos 3x = 0
Solution:
Given that y = x sin 3x ………(i)
On differentiating both sides w.r.t. x, we get
\(\frac{d y}{d x}=x \frac{d}{d x}(\sin 3 x)+\sin 3 x \frac{d}{d x}(x)\) [Using product rule]
\(\frac {dy}{dx}\) = x cos 3x . 3 + sin 3x
Again, differentiating on both sides w.r.t.x, we get
\(\frac{d^2 y}{d x^2}=3\left[x \frac{d}{d x} \cos 3 x+\cos 3 x \frac{d}{d x}(x)\right]+\frac{d}{d x}(\sin 3 x)\)
⇒ \(\frac{d^2 y}{d x^2}\) = 3[x(-sin 3x . 3) + cos 3x] + cos 3x . 3
⇒ \(\frac{d^2 y}{d x^2}\) = -9x sin 3x + 3 cos 3x + 3 cos 3x
⇒ \(\frac{d^2 y}{d x^2}\) = -9x sin 3x + 6 cos 3x
⇒ \(\frac{d^2 y}{d x^2}\) = -9y + 6 cos 3x = 0 [∵ from (i)]
⇒ \(\frac{d^2 y}{d x^2}\) + 9y – 6 cos 3x = 0
∴ The given function is a solution of the corresponding differential equation.
Question 7.
x2 = 2y2 log y; (x2 + y2) \(\frac{d y}{d x}\) – xy = 0
Solution:
Given that x2 = 2y2 log y …..(1)
On differentiating both sides w.r.t. x, we get
2x = \(2\left[y^2 \times \frac{1}{y} \frac{d y}{d x}+\log y \cdot 2 y \frac{d y}{d x}\right]\)
⇒ x = (y + 2y log y) \(\frac{d y}{d x}\)
On multiplying both sides by y, we get
xy = (y2 + 2y2 log y) \(\frac{d y}{d x}\)
⇒ xy = (y2 + x2) \(\frac{d y}{d x}\) [∵ from (i)]
⇒ (x2 + y2) \(\frac{d y}{d x}\) – xy = 0
Hence, the given function is a solution of the corresponding differential equation.
Question 8.
Find the general solution of the differential equation \(\frac{d y}{d x}+\sqrt{\frac{1-y^2}{1-x^2}}=0.\)
Solution:
Given differential equation is \(\frac{d y}{d x}+\sqrt{\frac{1-y^2}{1-x^2}}=0\)
⇒ \(\frac{d y}{d x}=-\frac{\sqrt{1-y^2}}{\sqrt{1-x^2}}\)
⇒ \(\frac{1}{\sqrt{1-y^2}} d y=-\frac{1}{\sqrt{1-x^2}} d x\)
Integrating both sides, we get
\(\int \frac{1}{\sqrt{1-y^2}} d y=-\int \frac{1}{\sqrt{1-x^2}} d x\)
⇒ sin-1 y = -sin-1 x + C
⇒ sin-1 x + sin-1 y = C
Which is the required general solution.
II.
Question 1.
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.
Solution:
The differential equation of the given curve is
sin x cos y dx + cos x sin y dy = 0
⇒ \(\frac{\sin x}{\cos x} d x+\frac{\sin y}{\cos y} d y=0\)
⇒ tan x dx + tan y dy = 0
On integrating both sides, we get
⇒ ∫tan x dx + ∫tan y dy = log C
⇒ log (sec x) + log (sec y) = log C
⇒ sec x . sec y = C …..(i)
The curve passes through the point (0, \(\frac{\pi}{4}\))
Put x = 0, y = \(\frac{\pi}{4}\), we get
sec 0 . sec \(\frac{\pi}{4}\) = C
⇒ C = √2
On putting the value of C in equation (1), we get
sec x . sec y = √2
⇒ \(\sec x \cdot \frac{1}{\cos y}=\sqrt{2}\)
⇒ cos y = \(\frac{\sec x}{\sqrt{2}}\)
Hence, the required equation of the curve is cos y = \(\frac{\sec x}{\sqrt{2}}\)
Question 2.
Find the particular solution of the differential equation (1 + e2x) dy + (1 + y2) ex dx = 0, given that y = 1 when x = 0.
Solution:
The given differential equation is
(1 + e2x) dy + (1 + y2) ex dx = 0
Separating the variables, we get
\(\frac{d y}{1+y^2}+\frac{e^x d x}{1+e^{2 x}}=0\)
Integrating both sides, we get
\(\int \frac{d y}{1+y^2}+\int \frac{e^x d x}{1+e^{2 x}}\) = C
Put t = ex
⇒ ex dx = dt
⇒ tan-1 y + \(\int \frac{d t}{1+t^2}\) = C
⇒ tan-1 y + tan-1 t = C
⇒ tan-1 y + tan-1 ex = C …..(1)
Now put x = 0, y = 1
∴ tan-1 (1) + tan-1 (e0) = C
⇒ \(\frac{\pi}{4}+\frac{\pi}{4}\) = C
⇒ C = \(\frac{\pi}{2}\)
On putting the value of C in equation (1), we get
\(\tan ^{-1} y+\tan ^{-1} e^x=\frac{\pi}{2}\)
Which is the required particular solution.
Question 3.
Solve the differential equation \(y e^{\frac{x}{y}} d x=\left(x e^{\frac{x}{y}}+y^2\right) d y(y \neq 0)\)
Solution:
Given differential equation is \(y e^{\frac{x}{y}} d x=\left(x e^{\frac{x}{y}}+y^2\right) d y(y \neq 0)\)
⇒ \(e^{x / y} \frac{d x}{d y}=\frac{x}{y} e^{x / y}+y\)
⇒ \(e^{x / y} \frac{d x}{d y}-\frac{x}{y} e^{x / y}=y\) ……(1)
Put x = vy
⇒ \(\frac{d x}{d y}=v+y \frac{d v}{d y}\)
The equation (1) becomes
\(e^v\left[v+y \frac{d v}{d y}\right]-v e^v=y\)
⇒ \(e^v y \frac{d v}{d y}\) = y
⇒ \(\mathrm{e}^{\mathrm{v}} \frac{\mathrm{dv}}{\mathrm{dy}}\)
⇒ ev dv = dy
On integrating both sides, we get
⇒ ∫ev dv = ∫1 dy
⇒ ev = y + C
⇒ \(e^{x / y}\) = y + C
Question 4.
Solve the differential equation \(\left[\frac{e^{-2 \sqrt{x}}}{\sqrt{x}}-\frac{y}{\sqrt{x}}\right] \frac{d x}{d y}=1(x \neq 0)\)
Solution:
The given differential equation is
\(\left(\frac{e^{-2 \sqrt{x}}}{\sqrt{x}}-\frac{y}{\sqrt{x}}\right) \frac{d x}{d y}=1\) ……(i)
⇒ \(\frac{d y}{d x}=\frac{e^{-2 \sqrt{x}}}{\sqrt{x}}-\frac{y}{\sqrt{x}}\)
⇒ \(\frac{d y}{d x}+\frac{1}{\sqrt{x}} y=\frac{e^{-2 \sqrt{x}}}{\sqrt{x}}\)
On comparing with \(\frac {dy}{dx}\) + Py = Q, we get
\(P=\frac{1}{\sqrt{x}} ; Q=\frac{e^{-2 \sqrt{x}}}{\sqrt{x}}\)
I.F = \(\mathrm{e}^{\int \mathrm{Pdx}}=\mathrm{e}^{\int \frac{1}{\sqrt{\mathrm{x}}} \mathrm{dx}}=\mathrm{e}^{2 \mathrm{x}^{1 / 2}}=\mathrm{e}^{2 \sqrt{\mathrm{x}}}\)
The general solution is y . IF = ∫Q × (I.F) dx + C
⇒ \(y e^{2 \sqrt{x}}=\int e^{2 \sqrt{x}} \cdot \frac{e^{-2 \sqrt{x}}}{\sqrt{x}} d x+C\)
⇒ \(y e^{2 \sqrt{x}}=\int \frac{1}{\sqrt{x}} d x+C\)
⇒ \(y e^{2 \sqrt{x}}=2 \sqrt{x}+C\)
Question 5.
Find a particular solution of the differential equation \(\frac {dy}{dx}\) + y cot x = 4x cosec x (x ≠ 0), given that y = 0 when x = \(\frac{\pi}{2}\).
Solution:
The given differential equation is
\(\frac {dy}{dx}\) + y cot x = 4x cosec x ……(i)
On comparing with \(\frac {dy}{dx}\) + Py = Q, we get
P = cot x, Q = 4x cosec x
IF = \(\mathrm{e}^{\int \cot \mathrm{xdx}}=\mathrm{e}^{\log |\sin \mathrm{x}|}\) = sin x
∴ The general solution is y . (I.F) = ∫Q.dx + C
⇒ y sin x = ∫4x cosec x sin x dx + C
⇒ y sinx = ∫4x dx + C
⇒ y sin x = 4 . \(\frac{x^2}{2}\) + C
⇒ y sin x = 2x2 + C …..(ii)
Put, x = \(\frac{\pi}{2}\) and y = 0
⇒ \(2\left(\frac{\pi}{2}\right)^2+C\) = 0
⇒ C = \(\frac{-\pi^2}{2}\)
On putting the value of C in equation (ii), we get
y sin x = \(2 x^2-\frac{\pi^2}{2}\)
III.
Question 1.
Prove that x2 – y2 = c(x2 + y2)2 is the general solution of differential equation (x3 – 3xy2) dx = (y3 – 3x2y) dy, where c is a parameter.
Solution:
Given differential equation is (x3 – 3xy2) dx = (y3 – 3x2y) dy
It can be rewritten as
Put y = vx
\(\frac{d y}{d x}=\frac{x^3-3 x y^2}{y^3-3 x^2 y}\) ……..(i)
This is a homogeneous equation
⇒ \(\frac{d y}{d x}=v+x \frac{d v}{d x}\)
Then, equation (i) becomes
\(v+x \frac{d v}{d x}=\frac{x^3-3 x(v x)^2}{(v x)^3-3 x^2(v x)}\)
⇒ \(v+x \frac{d v}{d x}=\frac{1-3 v^2}{v^3-3 v}\)
⇒ \(x \frac{d v}{d x}=\frac{1-3 v^2}{v^3-3 v}-v\)
⇒ \(x \frac{d v}{d x}=\frac{1-3 v^2-v^4+3 v^2}{v^3-3 v}\)
⇒ \(x \frac{d v}{d x}=\frac{1-v^4}{v^3-3 v}\)
⇒ \(\left(\frac{v^3-3 v}{1-v^4}\right) d v=\frac{d x}{x}\)
On integrating both sides, we get
⇒ \(\int\left(\frac{v^3-3 v}{1-v^4}\right) d v=\int \frac{d x}{x}\)
⇒ \(\int\left(\frac{v^3-3 v}{1-v^4}\right) d v\) = log x + log C …….(ii)
Now, \(\int\left(\frac{v^3-3 v}{1-v^4}\right) d v=\int \frac{v^3}{1-v^4} d v-3 \int \frac{v}{1-v^4} d v\) = I1 – 3I2 ……(iii)
Where, \(I_1=\int \frac{v^3}{1-v^4} d v\) and \(I_2=\int \frac{v}{1-v^4} d v\)
Put 1 – v4 = t
⇒ -4v3 = \(\frac {dt}{dv}\)
⇒ v3 dv = \(-\frac {dt}{4}\)
∴ \(I_1=\int \frac{-d t}{4 t}=\frac{-1}{4} \log t=\frac{-1}{4} \log \left(1-v^4\right)\)
and \(I_2=\int \frac{v d v}{1-v^4}=\int \frac{v d v}{1-\left(v^2\right)^2}\)
Put v2 = z
⇒ v dv = \(\frac {dz}{2}\)
⇒ \(I_2=\frac{1}{2} \int \frac{d z}{1-z^2}=\frac{1}{2 \times 2} \log \left|\frac{1+z}{1-z}\right|\)
= \(\frac{1}{4} \log \left|\frac{1+v^2}{1-v^2}\right|\) (∵ z = v2)
[∵ \(\int \frac{d x}{a^2-x^2}=\frac{1}{2 a} \log \left|\frac{a+x}{a-x}\right|\)]
On substituting the values of I1 and I2 in equation (iii), we get
\(\int\left(\frac{v^3-3 v}{1-v^4}\right) d v=\frac{-1}{4} \log \left(1-v^4\right)-\frac{3}{4} \log \left|\frac{1+v^2}{1-v^2}\right|\)
Equation (ii) becomes
⇒ \(\frac{-1}{4} \log \left(1-v^4\right)-\frac{3}{4} \log \left|\frac{1+v^2}{1-v^2}\right|\) = log x + log C
⇒ \(\frac{-1}{4} \log \left[\left(1-v^4\right)\left(\frac{1+v^2}{1-v^2}\right)^3\right]\) = log Cx
⇒ \(\frac{-1}{4} \log \left[\left(1-v^2\right)\left(1+v^2\right) \times \frac{\left(1+v^2\right)^3}{\left(1-v^2\right)^3}\right]\) = log Cx
⇒ \(\log \left[\frac{\left(1+v^2\right)^4}{\left(1-v^2\right)^2}\right]^{-1 / 4}\) = log Cx
⇒ \(\frac{\left(1+v^2\right)^4}{\left(1-v^2\right)^2}=(C x)^{-4}\)
⇒ \(\frac{\left(1+\frac{y^2}{x^2}\right)^4}{\left(1-\frac{y^2}{x^2}\right)^2}=\frac{1}{C^4 x^4}\)
⇒ \(\frac{\left(x^2+y^2\right)^4}{x^4\left(x^2-y^2\right)^2}=\frac{1}{C^4 x^4}\)
⇒ (x2 – y2) = C2 (x2 + y2)2
⇒ x2 – y2 = C(x2 + y2)2
Where C = C2
Hence, the given result is proved.
Question 2.
Show that the general solution of the differential equation \(\frac{d y}{d x}+\frac{y^2+y+1}{x^2+x+1}=0\) is given by (x + y + 1) = A(1 – x – y – 2xy), where A is a parameter.
Solution:
The given differential equation is
\(\frac{d y}{d x}+\frac{y^2+y+1}{x^2+x+1}=0\)
⇒ \(\frac{d y}{y^2+y+1}+\frac{d x}{x^2+x+1}=0\)
Integrating both sides, we get
\(\int \frac{d y}{y^2+y+1}+\int \frac{d x}{x^2+x+1}=C\)
⇒ \(\int \frac{d y}{y^2+y+1+\left(\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2}\) + \(\int \frac{d x}{x^2+x+1+\left(\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2}=C\)
⇒ \(\int \frac{\mathrm{dy}}{\left(\mathrm{y}+\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}+\int \frac{\mathrm{dx}}{\left(\mathrm{x}+\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}=\mathrm{C}\)
⇒ \(\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{y+\frac{1}{2}}{\sqrt{3} / 2}\right)+\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{x+\frac{1}{2}}{\sqrt{3} / 2}\right)=C\) [∵ \(\int \frac{1}{a^2+x^2} d x=\frac{1}{a} \tan ^{-1} \frac{x}{2}\)]
⇒ \(\tan ^{-1}\left(\frac{2 y+1}{\sqrt{3}}\right)+\tan ^{-1}\left(\frac{2 x+1}{\sqrt{3}}\right)=\frac{\sqrt{3} C}{2}=k\)
⇒ \(\tan ^{-1}\left[\frac{\frac{2 y+1}{\sqrt{3}}+\frac{2 x+1}{\sqrt{3}}}{1-\left(\frac{2 y+1}{\sqrt{3}}\right)\left(\frac{2 x+1}{\sqrt{3}}\right)}\right]=k\) [∵ \(\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)\)]
⇒ \(\tan ^{-1}\left[\frac{\frac{2 y+1+2 x+1}{\sqrt{3}}}{1-\left(\frac{4 x y+2 x+2 y+1}{3}\right)}\right]=k\)
⇒ \(\frac{2 \sqrt{3}(x+y+1)}{3-(4 x y+2 x+2 y+1)}\) = tan k
⇒ \(\frac{2 \sqrt{3}(x+y+1)}{2(1-x-y-2 x y)}\) = tan k
⇒ x + y + 1 = \(\frac{1}{\sqrt{3}}\) tan k (1 – x – y – 2xy)
Let A = \(\frac{1}{\sqrt{3}}\) tan k is an arbitrary constant.
⇒ x + y + 1 = A(1 – x – y – 2xy)
Hence, the given result is proved.
Question 3.
Find a particular solution of the differential equation (x – y)(dx + dy) = dx – dy, given that y = -1, when x = 0. (Hint: put x – y = t).
Solution:
Given differential equation is (x – y)(dx + dy) = dx – dy
⇒ dx + dy = \(\frac{d x-d y}{x-y}\)
On integrating both sides, we get
\(\int(d x+d y)=\int \frac{d x-d y}{x-y}+C\)
Let x – y = t
⇒ dx – dy = dt
∴ \(\int \mathrm{dx}+\mathrm{dy}-\mathrm{C}=\int \frac{\mathrm{dx}-\mathrm{dy}}{\mathrm{x}-\mathrm{y}}=\int \frac{\mathrm{dt}}{\mathrm{t}}\) = log t = log (x – y)
⇒ x + y = log |x – y| + C ……..(i)
It is given that when x = 0, y = -1
∴ 0 + (-1) = log (0 + 1) + C
⇒ C = -1
On substituting this value into equation (i), we get
x + y = log|x – y| – 1
⇒ log |x – y| = x + y + 1
Which is the required particular solution.
Question 4.
Find a particular solution of the differential equation \((\mathrm{x}+1) \frac{\mathrm{dy}}{\mathrm{dx}}=2 \mathrm{e}^{-\mathrm{y}}-1\), given that y = 0 when x = 0.
Solution:
The given differential equation is
\((\mathrm{x}+1) \frac{\mathrm{dy}}{\mathrm{dx}}=2 \mathrm{e}^{-\mathrm{y}}-1\)
On separating the variables
⇒ \(\frac{d y}{2 e^{-y}-1}=\frac{d x}{x+1}\)
⇒ \(\frac{e^y d y}{2-e^y}=\frac{d x}{x+1}\)
On integrating both sides, we get
\(\int \frac{e^y d y}{2-e^y}=\int \frac{d x}{x+1}\) …..(ii)
Put 2 – ey = t
⇒ -ey = \(\frac {dt}{dy}\)
⇒ ey dy = -dt
Then equation (ii) becomes
\(\int \frac{-\mathrm{dt}}{\mathrm{t}}=\int \frac{\mathrm{dx}}{\mathrm{x}+1}\)
⇒ -log |t| = log |x + 1| + log C
⇒ -log |2 – ey| = log |C(x + 1)|
⇒ \(\frac{1}{2-e^y}\) = C(x + 1) [∵ -log x = log x-1 = \(\frac {1}{x}\)]
⇒ 2 – ey = \(\frac{1}{C(x+1)}\) ….(iii)
Now, at x = 0 & y = 0
⇒ 2 – 1 = \(\frac {1}{C}\)
⇒ C = 1
On putting the value of C in equation (iii), we get
2 – ey = \(\frac{1}{(x+1)}\)
⇒ ey = 2 – \(\frac{1}{x+1}\)
⇒ \(e^y=\frac{2 x+2-1}{x+1}\)
⇒ \(e^y=\frac{2 x+1}{x+1}\)
⇒ y = \(\log \left|\frac{2 x+1}{x+1}\right|\), (x ≠ -1)
[∵ If loge x = m ⇒ em = x]
This is the required particular solution.