Answer to Question #132208 in Differential Equations for Rouhish Ray

Question

Solve: 2siny^2dx + xycos^2dy=0 ,
Given: y(2)=√π/2

Answer 1

As per the question ,

Given differential equation is

$2sin(y^2)dx+xycos(y^2)dy=0$

Using seperation of variable method,

xy$cos(y^2)$ $dy=-2sin(y^2)dx$

$\frac{ycos(y^2)}{sin(y^2)}dy=-\frac{2}{x}dx$

Integrating both the side $\int ycot(y^2)dy=-\int \frac{2}{x}dx$ ....(1)

let y$^2=t$

Diffrentiate with respect to x

2ydy=dt

ydy=$\frac{dt}{2}$

Substitute y $^2$ =t in equation 1

$\int \frac{Cot tdt}{2}=-2logx+logc$

$\frac{log|sin t|}{2}=logx^{-2}+logc$

$log|sin (y^2)|^{0.5}=logx^{-2}+logc$

$log|sin (y^2)|^{0.5}=logx^{-2}c$ { using loga+logb=logab}

$sin(y^2)^{0.5}=\frac{c}{x^2}$

$\sqrt{sin(y^2)}=\frac{c}{x^2}$ ....(2)

at Y(2)=$\sqrt{\frac{\Pi}{2}}$

$\sqrt{sin(\frac{\Pi}{2})}=\frac{c}{2^2}$

1=$\frac{c}{4}$

c=4

Putting the value of c in equation 2

$\sqrt{sin(y^2)}=\frac{4}{x^2}$

This is the required solution.