Solve the following initial value problem: `{xsin^2(y/x)-y}dx+x dy=0,\ y(1)=pi/4`

Explanation

Given differential equation is, `[x \sin ^{2}(\frac{y}{x})-y] d x+x d y=0`

`frac{d y}{d x}=\frac{y x \sin ^{2}(\frac{y}{x})}{x} \ldots` (i)

Let `F(x, y)=\frac{y-x \sin ^{2}(\frac{y}{x})}{x}`

Then `F(\lambda x, \lambda y)=\frac{\lambda y-\lambda x \sin ^{2} \frac{\lambda y}{\lambda x}}{\lambda x}=\lambda^{\prime} \frac{y-x \sin ^{2} \frac{y}{x}}{x}=\lambda^{\circ} F(x, y)`

Hence, differential equation (i) is homogeneous.

Now, let `y=v x` And`frac{d y}{d x}=v+x \frac{d v}{d x}`

Putting these value in (i), we get `\frac{d v}{d x}=\frac{v x-x \sin ^{2} \frac{v x}{x}}{x}`

`v+x \frac{d v}{d x}=\frac{x v-\sin ^{2} v}{x}`

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