Differential Equations of First Order and First Degree Solved Example Problems Exercise 2.9
Example. 1: Solve \(\frac{d y}{d x}+\frac{y}{x}=y^2 x \sin x, x>0\)
Solution.
Given equation is \(\frac{d y}{d x}+\frac{y}{x}=y^2 x \sin x\) ………………(1) is Bernoulli’s equation.
Multiplying (1) by \(y^{-2} \text {, we get : } y^{-2} \frac{2 y}{d x}+\frac{y}{x}=x \sin x\) ………………(2)
Let \( y^{-1}=u \Rightarrow(-1) y^{-2} \frac{d y}{d x}=\frac{d u}{d x} \Rightarrow y^{-2} \frac{d y}{d x}=-\frac{d u}{d x}\) ……………..(3)
(2) and (3) \(\Rightarrow-\frac{d u}{d x}+\frac{u}{x}=x \sin x \Rightarrow \frac{d u}{d x}-\frac{u}{x}=-x \sin x\) ………..(4)
(4) is a linear equation in u and x where \(\mathrm{P}=-\frac{1}{x}, \mathrm{Q}=-x \sin x\)
∴ \(\text { I.F. }=\exp \left(\int \mathrm{P} d x\right)=\exp \left(\int-\frac{1}{x} d x\right)=e^{-\log x}=e^{\log x^{-1}}=\frac{1}{x}\)
The general solution of (4) is I.F \(=\int \mathrm{Q} \text { (I.F) } d x+c \Rightarrow u(1 / x)=\int(-x \sin x)(1 / x) d x+c=\int(-\sin x) d x+c\)
⇒ \(u(1 / x)=\cos x+c \Rightarrow u=x \cos x+c x\) …………………..(5)
substitution u = 1/y in (5), the general solution of (1) is
∴ \(\text { 1/ } y=x \cos x+c x \Rightarrow x y \cos x+c x y=1\)
Differential Equations Of First Order And First Degree Exercise 2.9
Example. 2: Solve \(x \frac{d y}{d x}+y=y^2 \log x\)
Solution:
Given equation is \(x \frac{d y}{d x}+y=y^2 \log x\) ……………………….(1)
Dividing (1) by x and then multiplying with \(y^{-2}\)2 we get:
⇒ \(\frac{d y}{d x}+\frac{y}{x}=\frac{y^2 \log x}{x} \Rightarrow \frac{1}{y^2} \frac{d y}{d x}+\frac{1}{x} \cdot \frac{1}{y}=\frac{\log x}{x}\) …………………..(2)
Let \(u=\frac{1}{y} \Rightarrow \frac{d u}{d x}=-\frac{1}{y^2} \frac{d y}{d x} \Rightarrow \frac{1}{y^2} \frac{d y}{d x} \Rightarrow=-\frac{d u}{d x}\) ……………(3)
(2) and (3) \(\Rightarrow-\frac{d u}{d x}+\frac{1}{x} u=\frac{\log x}{x} \Rightarrow \frac{d u}{d x}-\frac{1}{x} u=\frac{-\log x}{x}\) ……………..(4)
(4) is a linear equation in and x where \(\mathrm{P}=-\frac{1}{x}, \mathrm{Q}=\frac{-\log x}{x}\)
Then \(\text { I.F. }=\exp \left(\int \mathrm{P} d x\right)=\exp \left(\int-\frac{1}{x} d x\right)=e^{-\log x}=e^{\log x^{-1}}=\frac{1}{x}\)
The General solution of (4) is (I.F.) = \(\int \mathrm{Q}(\text { I.F) } d x+c\)
⇒ \(u\left(\frac{1}{x}\right)=\int \frac{-\log x}{x} \cdot \frac{1}{x} d x+c=\int-\frac{1}{x^2} \log x d x+c\)
⇒ \(u\left(\frac{1}{x}\right)=\frac{1}{x} \log x-\int \frac{1}{x} \cdot \frac{1}{x} d x+c=\frac{1}{x} \log x+\frac{1}{x}+c\) ………………………….(5)
Putting \(u=\frac{1}{y}\) in (5), the general solution of(1) is
∴ \(\frac{1}{x y}=\frac{1}{x} \log x+\frac{1}{x}+c \Rightarrow 1=y \log x+y+c x y\)
Homogeneous Equations Solved Problems Exercise 2.9
Example. 3: Solve \(\frac{d y}{d x}+\frac{x y}{1-x^2}=x \sqrt{y}\)
Solution.
Given equation is \(\frac{d y}{d x}+\frac{x y}{1-x^2}=x \sqrt{y} \) ………………………(1)
Multiplying (1) by \(y^{-1 / 2} \Rightarrow y^{-1 / 2} \frac{d y}{d x}+\frac{x}{1-x^2} y^{1 / 2}=x\) ………………………(2)
Let \(y^{1 / 2}=u \Rightarrow \frac{1}{2} y^{-1 / 2} \frac{d y}{d x}=\frac{d u}{d x} \Rightarrow y^{-1 / 2} \frac{d y}{d x}=2 \frac{d u}{d x}\) ……………………(3)
(2) and (3) \(\Rightarrow 2 \frac{d u}{d x}+\frac{x}{1-x^2} u=x \Rightarrow \frac{d u}{d x}+\frac{x}{2\left(1-x^2\right)} u=\frac{x}{2}\)
where \(\mathrm{P}=\frac{x}{2\left(1-x^2\right)} \text { and } \mathrm{Q}=\frac{x}{2}\)
Now \(\text { I.F. }=\exp \left(\int \frac{x}{2\left(1-x^2\right)} d x\right)=\exp \left(\int \frac{-2 x}{-4\left(1-x^2\right)} d x\right)\)
= \(\exp \left[-\frac{1}{4} \log \left(1-x^2\right)\right]=\exp \left[\log \left(1-x^2\right)^{-1 / 4}\right]=\frac{1}{\left(1-x^2\right)^{1 / 4}}\)
The G. S. of (4) is \(u \cdot \frac{1}{\left(1-x^2\right)^{1 / 4}}=\int \frac{x}{2} \cdot \frac{1}{\left(1-x^2\right)^{1 / 4}} d x+c\)
⇒ \(\frac{u}{\left(1-x^2\right)^{1 / 4}}=-\frac{1}{4} \int t^{-1 / 4} d t+c \text { where } t=1-x^2\)
⇒ \(\frac{u}{\left(1-x^2\right)^{1 / 4}}=-\frac{1}{4} \frac{t^{3 / 4}}{(3 / 4)}+c=-\frac{1}{3}\left(1-x^2\right)^{3 / 4}+c\)
∴ The G.S of (1) is \(\frac{\sqrt{y}}{\left(1-x^2\right)^{1 / 4}}=-\frac{1}{3}\left(1-x^2\right)^{3 / 4}+c \Rightarrow 3 \sqrt{y}+\left(1-x^2\right)=3 c\left(1-x^2\right)^{1 / 4}\)
Methods To Find Integrating Factors For Exercise 2.9
Example.4 Solve \(\frac{d y}{d x}\left(x^2 y^3+x y\right)=1\)
Solution.
Given equation is \(\frac{d y}{d x}\left(x^2 y^3+x y\right)=1 \Rightarrow \frac{d x}{d y}=x^2 y^3+x y \Rightarrow \frac{d x}{d y}-x y=x^2 y^3\) ……………..(1)
(1) is Bernoulli’s equation in x. Multiplying (1) by \(x^{-2} \Rightarrow x^{-2} \frac{d x}{d y}-x^{-1} y=y^3\) ……….(2)
Put \(x^{-1}=u \Rightarrow-1. x^{-2} \frac{d x}{d y}=\frac{d u}{d y} \Rightarrow x^{-2} \frac{d x}{d y}=-\frac{d u}{d y}\) ……………..(3)
(2) and (3) \(\Rightarrow-\frac{d u}{d y}-u y=y^3 \Rightarrow \frac{d u}{d y}+u y=-y^3\) ………….(4)
(4) is a linear equation in u and y where \(\mathrm{P}=y, \mathrm{Q}=-y^3 \text {. Now I.F. }=\exp \left(\int y d y\right)=e^{y^2 / 2}\)
G. S. of (4) is \(u e^{y^2 / 2}=\int-y^3 e^{y^2 / 2} d y+c=-2 \int t e^t d t+c \text { where } t=y^2 / 2\)
⇒ \(u e^{y^2 / 2}=-2 e^t(t-1)+c\) Substituting u and t
The G.S. of (1) is \(\frac{1}{x} e^{y^2 / 2}=-2 e^{y^2 / 2}\left(\frac{y^2}{2}-1\right)+c \Rightarrow x\left(2-y^2\right)-c x e^{-y^2 / 2}=1\)
Solved Example Problems From Exercise 2.9 In Differential Equations
Example.5. Solve \(\frac{d y}{d x}=\frac{x^2+y^2+1}{2 x y}\)
Solution.
Given equation
\(\frac{d y}{d x}=\frac{x^2+y^2+1}{2 x y}\)Given equation can be written as \(2 x y \frac{d y}{d x}=x^2+y^2+1 \Rightarrow 2 x y \frac{d y}{d x}-y^2=x^2+1\)
⇒ \(2 y \frac{d y}{d x}-\left(\frac{1}{x}\right) y^2=\frac{x^2+1}{x} . \text { Put } y^2=z \Rightarrow 2 y \frac{d y}{d x}=\frac{d z}{d x}\)
∴ \(\frac{d z}{d x}-\frac{1}{x} z=\frac{x^2+1}{x}\) This is a linear equation in z
Where \(\mathrm{P}=-1 / x \text { I.F. }=e^{\int(-1 / x) d x}=e^{-\log x}=e^{\log x^{-1}}=x^{-1}=1 / x\)
⇒ \(e^{\int(-1 / x) d x}=e^{-\log x}=e^{\log x^{-1}}=x^{-1}=1 / x\)
∴ \(\text { G.S. is } z(1 / x)=\int \frac{1}{x} \cdot \frac{x^2+1}{x} d x+c=\int \frac{x^2+1}{x^2} d x+c\)
⇒ \(\frac{z}{x}=\int\left(1+\frac{1}{x^2}\right) d x+c=\int d x+\int \frac{1}{x^2} d x+c=x-\frac{1}{x}+c \Rightarrow z=x^2-1+c x\)
∴ The G.S. of the given equation is \(y^2=x^2-1+c x \Rightarrow y^2-x^2+1=c x\)
Solutions For Exercise 2.9 First-Order Homogeneous Equations
Example. 6. Solve \(x y-\frac{d y}{d x}=y^3 e^{-x^2}\)
Solution.
Given equation
\(x y-\frac{d y}{d x}=y^3 e^{-x^2}\)Given equation can be written as \(\frac{d y}{d x}-x y=-y^3 e^{-x^2}\)
⇒ \(\frac{1}{y^3} \frac{d y}{d x}-\frac{1}{y^2} x=-e^{-x^2} \text {. Put } \frac{1}{y^2}=z \Rightarrow-\frac{2}{y^3} \frac{d y}{d x}=\frac{d z}{d x}\)
⇒ \(-\frac{1}{2} \frac{d z}{d x}-x z=-e^{-x^2} \Rightarrow \frac{d z}{d x}+2 x z=2 e^{-x^2}\)
This is a linear equation in z where \(\mathrm{P}=2 x . \text { I.F. }=e^{\int 2 x d x}=e^{x^2}\)
∴ G.S. is \(z e^{x^2}=2 \int e^{-x^2} \cdot e^{x^2} d x+c=2 \int d x+c=2 x+c\)
∴ G.S. of the given equation is \(\left(1 / y^2\right) e^{x^2}=2 x+c\)
Examples Of Integrating Factors In Homogeneous Equations Exercise 2.9
Example.7. Solve \(\frac{d y}{d x}=2 y \tan x+y^2 \tan ^2 x\)
Solution.
Given equation
\(\frac{d y}{d x}=2 y \tan x+y^2 \tan ^2 x\)The given equation can be written as \(\frac{d y}{d x}-(2 \tan x) y=y^2 \tan ^2 x \Rightarrow \frac{1}{y^2} \frac{d y}{d x}-(2 \tan x) \frac{1}{y}=\tan ^2 x\)
Let \(-\frac{1}{y}=z \Rightarrow \frac{1}{y^2} \frac{d y}{d x}=\frac{d z}{d x}\)
Then the above equation becomes: \(\frac{d z}{d x}+(2 \tan x) z=\tan ^2 x\)
This is a linear equation in z where P = 2 tan x.
Now \(\text { I.F. }=e^{\int 2 \tan x d x}=e^{2 \log \sec x}=e^{\log \sec ^2 x}=\sec ^2 x\).
∴ G.S. is \(z \sec ^2 x=\int \tan ^2 x \sec ^2 x d x-c\)
⇒ \(z \sec ^2 x=\int \tan ^2 x d(\tan x)-c=\left(\tan ^3 x\right) / 3-c\)
∴ G.S of the G.E is \(-(1 / y) \sec ^2 x=(1 / 3) \tan ^3 x-c y^{-1} \sec ^2 x=c-(1 / 3) \tan ^3 x\)