Chapter 7.3, Problem 14E

To determine

To find : the solution of the differential equation.

Answer to Problem 14E

The solution of differential equation is $y+\ln \left|y\right|\text{=}\sin x-x\cos x\text{+1}$ .

Explanation of Solution

Given information :

The differential equation is $y\text{'}=\frac{xy\sin x}{y+1}\text{, }y\left(0\right)=1$ .

Calculation : here the equation is separable, so write the equation in the terms of differentials and then integrate both sides,

  $\begin{array}{l}y\text{'}=\frac{xy\sin x}{y+1}\\ \frac{dy}{dx}=\frac{xy\sin x}{y+1}\text{ [}y\text{'}=\frac{dy}{dx}]\\ \frac{y+1}{y}dy\text{=}x\sin xdx\text{ [separate the variables]}\\ \left(1+\frac{1}{y}\right)dy=x\sin xdx\text{ [simplify]}\\ {\displaystyle \int \left(1+\frac{1}{y}\right)dy=}{\displaystyle \int x\sin xdx\text{ [integrate both side]}}\\ y+\ln \left|y\right|\text{=}\sin x-x\cos x\text{+}c\text{ [simplify integrate] }\end{array}$

Here $c$ is a arbitrary constant.

Initial condition is $y\left(0\right)=1$ , now substitute $x=0\text{ and }y=1$ ,

  $\begin{array}{l}y+\ln \left|y\right|\text{=}\sin x-x\cos x\text{+}c\\ 1+\ln \left|1\right|=\sin 0-0·\cos 0+c\text{ [substitute]}\\ \text{1+0=0}-0+c\text{ [ln|1|=0,sin0=0]}\\ 1=c\text{ [add]}\end{array}$

So the solution of equation is $y+\ln \left|y\right|\text{=}\sin x-x\cos x\text{+1}$

Hence, the solution of differential equation is $y+\ln \left|y\right|\text{=}\sin x-x\cos x\text{+1}$ .