Solve the following differential equation: `x^2dy+(x^2-x y+y^2)dx=0`

Explanation

It is given that

`x^{2} d y+(x^{2}-x y+y^{2}) d y=0 `

`x^{2} d y=(x y-x^{2}-y^{2}) d y `

`frac{d y}{d x}=\frac{x y-x^{2}-y^{2}}{x^{2}}`

This is a homogeneous differential equation.

Putting `y=v x` and `\frac{d y}{d x}=v+x \frac{d v}{d x}`, we get

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

` v+x \frac{d v}{d x}=v-1 `

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