Practicing the AP Inter 2nd Year Maths Study Material and AP Inter 2nd Year Maths Chapter 9 Differential Equations Solutions Exercise 9e will help students to clear their doubts quickly.
Inter 2nd Year Maths Differential Equations Solutions Exercise 9e
Differential Equations Exercise 9e Solutions
I.
Question 1.
Find the general solution of \(\frac {dy}{dx}\) + 2y = sin x.
Solution:
Given D.E is \(\frac {dy}{dx}\) + 2y = sin x
Compare with \(\frac {dy}{dx}\) + Py = Q
Here P = 2, Q = sin x
IF = \(\mathrm{e}^{\int \mathrm{p} \mathrm{dx}}=\mathrm{e}^{\int 2 \mathrm{dx}}=\mathrm{e}^{2 \mathrm{x}}\)
∴ General solution is y(I.F) = ∫Q (I.F) dx + C
y e2x = ∫e2x sin x dx + C ….(1)
Let I = ∫e2x sin x dx
= \(\sin x \int e^{2 x} d x-\left[\int e^{2 x} d x \frac{d}{d x}(\sin ) \int e^{2 x} d x\right] d x\)
= \(\frac{e^{2 x}}{2} \sin x-\frac{1}{2}\left[\int \cos x e^{2 x} d x\right]\)
= \(\frac{e^{2 x}}{2} \sin x-\frac{1}{2}\left[\cos x \int e^{2 x} d x-\int\left(\frac{d}{d x}(\cos x) \int e^{2 x} d x\right) d x\right]\)
= \(\frac{e^{2 x}}{2} \sin x-\frac{1}{2}\left[\frac{e^{2 x}}{2} \cos x+\int \sin x \frac{e^{2 x}}{2} d x\right]\)
= \(\frac{e^{2 x}}{2} \sin x-\frac{e^{2 x}}{4} \cos x-\frac{1}{4} I\)
\(\frac{5}{4} I=\frac{e^{2 x}}{4}(2 \sin x-\cos x)\)
I = \(\frac{e^{2 x}}{5}(2 \sin x-\cos x)\)
Put the value of I into (1),
⇒ y e2x = \(\frac{e^{2 x}}{2^2+1}\)(2 sin x – cos x) + C
⇒ y = \(\frac {1}{5}\)(2 sin x – cos x) + C e-2x
Question 2.
Find the general solution of \(\frac {dy}{dx}\) + 3y = e-2x.
Solution:
Given D.E is \(\frac {dy}{dx}\) + 3y = e-2x
Compare with \(\frac {dy}{dx}\) + Py = Q
Here P = 3, Q = e-2x
IF = \(\mathrm{e}^{\int \mathrm{Pdx}}=\mathrm{e}^{\int 3 \mathrm{dx}}=\mathrm{e}^{3 \mathrm{x}}\)
∴ General solution is y(I.F) = ∫Q (I.F) dx + C
⇒ y e3x = ∫e-2x e3x dx + C
⇒ y e3x = ∫ex dx + C
⇒ y e3x = ex + C
⇒ y = e-2x + C e-3x
Question 3.
Find the general solution of \(\frac{d y}{d x}+\frac{y}{x}=x^2\)
Solution:
Given D.E is \(\frac{d y}{d x}+\frac{y}{x}=x^2\)
Compare with \(\frac {dy}{dx}\) + Py = Q
Here P = \(\frac {1}{x}\), Q = x2
IF = \(\mathrm{e}^{\int \mathrm{Pdx}}=\mathrm{e}^{\int \frac{1}{\mathrm{x}} \mathrm{dx}}=\mathrm{e}^{\log \mathrm{x}}\) = x
∴ General solution is y(I.F) = ∫Q I.F dx + C
⇒ yx = ∫x2 . x dx + C
⇒ yx = ∫x3 dx + C
⇒ xy = \(\frac{x^4}{4}\) + C
Question 4.
Find the general solution of \(\frac {dy}{dx}\) + (sec x)y = tan x (0 ≤ x ≤ \(\frac{\pi}{2}\))
Solution:
Given D.E is \(\frac {dy}{dx}\) + (sec x)y = tan x
Compare with \(\frac {dy}{dx}\) + Py = Q
Here P = sec x, Q = tan x
I.F = \(\mathrm{e}^{\int \mathrm{Pdx}}=\mathrm{e}^{\int \sec \mathrm{x}}=\mathrm{e}^{\log |\sec \mathrm{x}+\tan \mathrm{x}|}\) = sec x + tan x
∴ General solution is y(I.F) = ∫Q (I.F) dx + C
⇒ y(sec x + tan x) = ∫tan x (sec x + tan x) dx + C
= ∫tan x sec x dx + ∫tan2x dx + C
= sec x + ∫sec2x dx – ∫dx + C
⇒ y(sec x + tan x) = sec x + tan x – x + C
Question 5.
Find the general solution of \(\cos ^2 x \frac{d y}{d x}+y=\tan x\left(0 \leq x<\frac{\pi}{2}\right)\).
Solution:
Given D.E is \(\cos ^2 x \frac{d y}{d x}+y=\tan x\)
\(\frac {dy}{dx}\) + y sec2x = tan x sec2x
Compare with \(\frac {dy}{dx}\) + Py = Q
Here P = sec2x, Q = tan x sec2x
IF = \(\mathrm{e}^{\int \mathrm{Pdx}}=\mathrm{e}^{\int \sec ^2 \mathrm{xdx}}=\mathrm{e}^{\tan \mathrm{x}}\)
∴ General solution is y(I.F) = ∫Q I.F dx + C
⇒ \(y e^{\tan x}=\int \tan x \sec ^2 x e^{\tan x} d x+C\)
Put tan x = t
⇒ sec2x dx = dt
∴ \(y^{\tan x}=\int t e^t d t+C\) = (t – 1) et + C
\(y e^{\tan x}=(\tan x-1) e^{\tan x}+C\)
\(\mathrm{y}=(\tan \mathrm{x}-1)+\mathrm{Ce}^{-\tan \mathrm{x}}\)
Question 6.
Find the general solution of x \(\frac {dy}{dx}\) + 2y = x2 log x.
Solution:
Given D.E is x \(\frac {dy}{dx}\) + 2y = x2 log x
⇒ \(\frac{d y}{d x}+\frac{2 y}{x}\) = x log x
Compare with \(\frac {dy}{dx}\) + Py = Q
Here P = \(\frac {2}{x}\), Q = x log x
IF = \(\mathrm{e}^{\int \mathrm{Pdx}}=\mathrm{e}^{\int \frac{2}{\mathrm{x}} \mathrm{dx}}=\mathrm{e}^{2 \log \mathrm{x}}=\mathrm{x}^2\)
∴ General solution is y(I.F) = ∫Q (I.F) dx + C
⇒ y x2 = ∫x log x . x2 dx + C
⇒ x2 y = ∫x3 log x dx + C
⇒ x2 y = \(\log x \int x^3 d x-\int\left(\frac{d}{d x}\right)(\log x)\left(\int x^3 d x\right) d x+C\)
⇒ x2 y = \(\frac{x^4}{4} \log x-\int \frac{1}{x} \frac{x^4}{4} d x+C\)
⇒ x2 y = \(\frac{x^4}{4} \log x-\frac{1}{4} \int x^3 d x+C\)
⇒ x2 y = \(\frac{x^4}{4} \log x-\frac{x^4}{16}+C\)
⇒ y = \(\frac{x^2}{16}(4 \log x-1)+C x^{-2}\)
Question 7.
Find the general solution of \(x \frac{d y}{d x}+y=\frac{2}{x} \log x\)
Solution:
Given D.E is \(x \frac{d y}{d x}+y=\frac{2}{x} \log x\)
⇒ \(\frac{d y}{d x}+\frac{y}{x \log x}=\frac{2}{x^2}\)
Compare with \(\frac {dy}{dx}\) + Py = Q
Here P = \(\frac{1}{x \log x}\), Q = \(\frac{2}{x^2}\)
I.F = \(\mathrm{e}^{\int \mathrm{pdx}}=\mathrm{e}^{\int \frac{1}{\mathrm{x} \log \mathrm{x}} \mathrm{dx}}=\mathrm{e}^{\log (\log \mathrm{x})}=\log \mathrm{x}\)
∴ General solution is y(I.F) = ∫Q (I.F) dx + C
y log x = \(\int \frac{2}{x^2} \log x \cdot d x\) + C
⇒ y log x = \(2 \int \log x \frac{1}{x^2} d x+C\)
= \(2\left(\log x \int \frac{1}{x^2} d x-\int \frac{1}{x}\left(\int \frac{1}{x^2} d x\right) d x\right)+C\)
= \(2\left(\frac{-\log x}{x}+\int \frac{1}{x^2} d x\right)+C\)
= \(2\left(\frac{-\log x}{x}-\frac{1}{x}\right)+C\)
∴ y log x = \(-\frac {2}{x}\)(1 + log x) + C
Question 8.
Find the general solution of (1 + x2) dy + 2xy dx = cot x dx (x ≠ 0)
Solution:
Given D.E is (1 +x2) dy + 2xy dx = cot x dx
⇒ (1 + x2) dy = (cot x – 2xy) dx
⇒ \(\frac{d y}{d x}+\frac{\cot x-2 x y}{1+x^2} \Rightarrow \frac{d y}{d x}+\frac{2 x y}{1+x^2}=\frac{\cot x}{1+x^2}\)
Compare with \(\frac {dy}{dx}\) + Py = Q
Here P = \(\frac{2 x}{1+x^2}\), Q = \(\frac{\cot x}{1+x^2}\)
I.F = \(\mathrm{e}^{\int \mathrm{Pdx}}=\mathrm{e}^{\int \frac{2 \mathrm{x}}{1+\mathrm{x}^2} \mathrm{dx}}=\mathrm{e}^{\log \left(1+\mathrm{x}^2\right)}\) = 1 + x2
∴ General solution is (I.F)y = ∫Q (I.F) dx + C
⇒ (1 + x2) y = \(\int \frac{\cot x}{1+x^2}\left(1+x^2\right) d x\) + C
⇒ (1 + x2) y = ∫ cot x dx + C
⇒ y(1 + x2) = log|sin x| + C
⇒ y = \(\frac{\log |\sin x|}{1+x^2}+\frac{C}{1+x^2}\)
Question 9.
Find the general solution of x \(\frac {dy}{dx}\) + y – x + xy cot x = 0 (x ≠ 0)
Solution:
Given D.E is x \(\frac {dy}{dx}\) + y – x + xy cot x = 0
⇒ \(\frac{d y}{d x}+\left(\frac{1}{x}+\cot x\right) y=1\)
Compare with \(\frac {dy}{dx}\) + Py = Q
Here P = \(\frac {1}{x}\) + cot x; Q = 1
I.F = \(\mathrm{e}^{\int \mathrm{Pdx}}=\mathrm{e}^{\int\left(\frac{1}{\mathrm{x}}+\cot \mathrm{x}\right) \mathrm{dx}}\) = \(\mathrm{e}^{\log x+\log \sin x}=\mathrm{e}^{\log x \sin x}\) = x sin x
∴ General solution is y(I.F) = ∫Q (I.F) dx + C
⇒ (x sin x) y = ∫x sin x dx + C
⇒ (x sin x) y = -x cos x + ∫cos x dx + C
⇒ (x sin x) y = -x cos x + sin x + C
⇒ y = \(\frac{-x \cos x}{x \sin x}+\frac{\sin x}{x \sin x}+\frac{C}{x \sin x}\)
⇒ y = \(\frac{1}{x}-\cot x+\frac{C}{x \sin x}\)
Question 10.
Find the general solution of (x + y) \(\frac {dy}{dx}\) = 1.
Solution:
Given D.E is (x + y) \(\frac {dy}{dx}\) = 1
\(\frac {dx}{dy}\) = x + y
\(\frac {dx}{dy}\) – x = y
Compare with \(\frac {dx}{dy}\) + Px = Q
Here P = -1; Q = y
I.F = \(\mathrm{e}^{\int \mathrm{Pdy}}=\mathrm{e}^{\int-1 \mathrm{dy}}=\mathrm{e}^{-\mathrm{y}}\)
∴ General solution is (I.F)x = ∫Q (I.F) dy + C
⇒ x e-y = ∫y e-y dy + C
⇒ x e-y = -y e-y + ∫e-y dy + C
⇒ x e-y = -y e-y – e-y + C
⇒ x = -y – 1 + C e-y
⇒ x + y + 1 = C ey
Question 11.
Find the general solution of y dx + (x – y2) dy = 0.
Solution:
Given y dx + (x – y2) dy = 0
⇒ \(\frac{d x}{d y}+\frac{x}{y}=y\)
Compare with \(\frac {dx}{dy}\) + Px = Q
Here P = \(\frac {1}{y}\); Q = y
I.F = \(\mathrm{e}^{\int \mathrm{P} \mathrm{dy}}=\mathrm{e}^{\int \frac{1}{\mathrm{y}} \mathrm{dy}}=\mathrm{e}^{\log \mathrm{y}}\) = y
∴ General solution is (I.F)x = ∫Q (I.F) dy + C
⇒ xy = ∫y . y dy + C
⇒ xy = ∫y2 dy + C
⇒ xy = \(\frac{y^3}{3}\) + C
⇒ x = \(\frac{y^2}{3}+\frac{c}{y}\)
Question 12.
Find the general solution of (x + 3y2) \(\frac {dy}{dx}\) = y (y > 0).
Solution:
Given D.E is (x + 3y2) \(\frac {dy}{dx}\) = y
⇒ y \(\frac {dx}{dy}\) = x + 3y2
⇒ \(\frac{d x}{d y}=\frac{x}{y}+3 y\) = 3y
⇒ \(\frac{d x}{d y}-\frac{x}{y}=3 y\)
Compare with \(\frac {dx}{dy}\) + Px = Q
Here P = \(-\frac {1}{y}\); Q = 3y
I.F = \(e^{\int P d y}=e^{\int-\frac{1}{y} d y}=e^{-\log y}=y^{-1}=\frac{1}{y}\)
∴ General solution is (I.F)x = ∫Q (I.F) dy + C
⇒ \(\frac{x}{y}=\int 3 y \cdot \frac{1}{y} d y+C\)
⇒ \(\frac {x}{y}\) = 3y + C
⇒ x = 3y2 + Cy
II.
Question 1.
Find the particular solution of \(\frac {dy}{dx}\) + 2y tan x = sin x; y = 0 when x = \(\frac{\pi}{3}\)
Solution:
Given \(\frac {dy}{dx}\) + 2y tan x = sin x
Compare with \(\frac {dy}{dx}\) + Px = Q
Here P= 2 tan x; Q = sin x
I.F = \(\mathrm{e}^{\int \mathrm{Pdx}}=\mathrm{e}^{\int 2 \tan \mathrm{xdx}}\)
= \(\mathrm{e}^{2 \log |\sec \mathrm{x}|}\)
= \(e^{\log \sec ^2 x}\)
= sec2x
∴ General solution is (I.F)y = ∫Q (I.F) dx + C
y sec2x = ∫sin x sec2 x dx + C = ∫tan x sec x dx + c
⇒ y sec2x = sec x + C …..(1)
Put y = 0 & x = \(\frac{\pi}{3}\) then
0 = sec \(\frac{\pi}{3}\) + C
⇒ C + 2 = 0
⇒ C = -2
Substitute ‘C’ value in equation (1)
⇒ y sec2x = sec x – 2
⇒ y = cos x – 2 cos2x
Which is the required solution.
Question 2.
Find the particular solution of \(\left(1+x^2\right) \frac{d y}{d x}+2 x y=\frac{1}{1+x^2}\); y = 0 when x = 1.
Solution:
Given D.E is \(\left(1+x^2\right) \frac{d y}{d x}+2 x y=\frac{1}{1+x^2}\)
⇒ \(\frac{d y}{d x}+\frac{2 x y}{1+x^2}=\frac{1}{\left(1+x^2\right)^2}\)
Compare with \(\frac {dy}{dx}\) + Px = Q
Here P = \(\frac{2 x}{1+x^2}\), Q = \(\frac{1}{\left(1+x^2\right)^2}\)
I.F = \(\mathrm{e}^{\int \mathrm{Pdx}}=\mathrm{e}^{\int \frac{2 \mathrm{x}}{1+\mathrm{x}^2}} \mathrm{dx}=\mathrm{e}^{\log (1+\mathrm{x})^2}=1+\mathrm{x}^2\)
∴ General solution is (I.F)y = ∫Q (I.F) dx + C
(1 + x2)y = \(\int \frac{1}{\left(1+x^2\right)^2}\left(1+x^2\right) d x+C\) = \(\int \frac{1}{1+x^2} d x+C\)
⇒ (1 + x2)y = tan-1x + C …….(1)
Put y = 0 & x = 1 then
0 = tan-1(1) + C
⇒ C = \(-\frac{\pi}{4}\)
Substitute ‘C’ value in equation (1)
y(1 + x2) = \(\tan ^{-1} x-\frac{\pi}{4}\)
Which is the required solution.
Question 3.
Find the particular solution of \(\frac {dy}{dx}\) – 3y cot x = sin 2x; y = 2 when x = \(\frac{\pi}{2}\)
Solution:
Given D.E is \(\frac {dy}{dx}\) – 3y cot x = sin 2x
Compare with \(\frac {dy}{dx}\) + Py = Q
Here P = -3 cot x; Q = sin 2x
I.F = \(\mathrm{e}^{\int \mathrm{Pdx}}=\mathrm{e}^{\int-3 \cot \mathrm{xdx}}=\mathrm{e}^{-3 \log |\sin \mathrm{x}|}\) = \(e^{\log (\sin x)^{-3}}=\frac{1}{\sin ^3 x}\)
∴ General solution is (I.F)y = ∫Q (I.F) dx + C
⇒ \(y \frac{1}{\sin ^3 x}=\int \frac{1}{\sin ^3 x} \sin 2 x d x+C\)
⇒ \(\frac{y}{\sin ^3 x}=\int \frac{2 \sin x \cos x}{\sin ^3 x} d x+C\)
⇒ \(\frac{y}{\sin ^3 x}\) = 2 ∫cot x cosec x dx + C
⇒ \(\frac{y}{\sin ^3 x}\) = -2 cosec x + C [∵ ∫cot x cosec x dx = -cosec x]
⇒ y = -2 sin3x cosec x + C sin3x
⇒ y = -2 sin2x + C sin3x ……….(1)
Put y = 2 and x = \(\frac{\pi}{2}\) then
2 = \(-2 \sin ^2 \frac{\pi}{2}+C \sin ^3 \frac{\pi}{2}\)
⇒ 2 = -2 + C
⇒ C = 4
Substitute ‘C’ value in equation (1)
y = -2 sin2x + 4 sin3x
Which is the required solution.
Question 4.
Find the equation of a curve passing through the origin, given that the slope of the tangent to the curve at any point (x, y) is equal to the sum of the coordinates of the point.
Solution:
Given DE is \(\frac {dy}{dx}\) = x + y
\(\frac {dy}{dx}\) – y = x
Compare with \(\frac {dy}{dx}\) + Py = Q
Here P = -1; Q = x
I.F = \(\mathrm{e}^{\int \mathrm{Pdx}}=\mathrm{e}^{-\int 1 \mathrm{dx}}=\mathrm{e}^{-\mathrm{x}}\)
∴ General solution is (I.F)y = ∫Q (I.F) dx + C
y e-x = ∫x e-x dx + C
y e-x = -x e-x – e-x + C
Given x = 0, y = 0 then C = 1
⇒ y e-x = -x e-x – e-x + 1
⇒ y = -x – 1 + ex
⇒ x + y + 1 = ex
Which is the required solution.
Question 5.
Find the equation of a curve passing through the point (0, 2) given that the sum of the coordinates of any point on the curve exceeds the magnitude of the slope of the tangent to the curve at that point by 5.
Solution:
Given DE is \(\frac {dy}{dx}\) + 5 = x + y
⇒ \(\frac {dy}{dx}\) – y = x – 5
Compare with \(\frac {dy}{dx}\) + Py = Q
Here P = -1; Q = x – 5
I.F = \(\mathrm{e}^{\int \mathrm{Pdx}}=\mathrm{e}^{-\int 1 \mathrm{dx}}=\mathrm{e}^{-\mathrm{x}}\)
∴ General solution is (I.F)y = ∫Q I.F dx + C
y e-x = ∫(x – 5) e-x dx + C
y e-x = -(x – 5) e-x – ∫-e-x dx + C
y e-x = (5 – x) e-x – e-x + C …….(1)
Equation (1) passes through (0, 2), then
2 = 5 – 1 + C
⇒ C = -2
Substitute ‘C’ value in equation (1), then
y e-x = (5 – x) e-x – e-x – 2
⇒ y = 4 – x – 2ex
Which is the required solution.