Question

Solve the Differential equation (x^2 + y^2) dx + (x^2 - xy) dy = 0

Answer 1

`(y)/(x)-2ln((y)/(x)+1)=lnx+C`

Explanation:

Given differential equation,

`(x^2 + y^2) dx + (x^2 - xy) dy = 0`

`\implies (dy)/(dx)=-(x^2 + y^2)/(x^2 - xy)----(1)`

Let y = vx

Differentiating with respect to x,

`(dy)/(dx)=v+x(dv)/(dx)`

From equation (1),

`v+x(dv)/(dx)=-(x^2 + (vx)^2)/(x^2 - x(vx))`

`v+x(dv)/(dx)=-(x^2 + v^2x^2)/(x^2 - vx^2)`

`v+x(dv)/(dx)=-(1 + v^2)/(1 - v)`

`v+x(dv)/(dx)=(1 + v^2)/(v-1)`

`x(dv)/(dx)=(1 + v^2)/(v-1)-v`

`x(dv)/(dx)=(1 + v^2-v^2+v)/(v-1)`

`x(dv)/(dx)=(v+1)/(v-1)`

`(v-1)/(v+1)dv=(1)/(x)dx`

Integrating both sides,

`\int{(v-1)/(v+1)}dv=\int{(1)/(x)}dx`

`\int{(v-1+1-1)/(v+1)}dv=lnx + C`

`\int{1-(2)/(v+1)}dv=lnx + C`

`v-2ln(v+1)=lnx+C`

Now, y = vx ⇒ v = y/x

`\implies (y)/(x)-2ln((y)/(x)+1)=lnx+C`