The solution of the equation $(x^2+xy)dy=(x^2+y^2)dx$is

The correct option is C $logx=(x-y)+\frac{y}{x}+c$

Given: $(x^2+xy)dy=(x^2+y^2)dx$

To find: Solution of the eq

Now,

$(x^2+xy)dy=(x^2+y^2)dx$

divide $x^2$ bo both side.

$\therefore (1+\frac{y}{x})dy=(1+\frac{y^2}{x^2})dx$

or , $(1+\frac{y}{x})\frac{dy}{dx}=(1+\frac{y^2}{x^2})$

Let, $\frac{y}{x}=v\Rightarrow y=xv$

or $\frac{dy}{dx}=v+x\frac{dv}{dx}$

$\therefore (1+v)(v+\frac{xdv}{dx})=(1+v^2)$

or, $v+x\frac{dv}{dx}=\frac{1+v^2}{1+v}-v$

or, $x\frac{dv}{dx}=\frac{1+v^2-v-v^2}{1+v}$

or $x\frac{dv}{dx}=\frac{1-v}{1+v}$

Integrating both side

$\therefore \int \left(\frac{1+v}{1-v}\right)dv=\int \frac{dx}{x}$

or, $\int (\frac{2}{1-v}-1)dv=\int \frac{dx}{x}$

or $\int \frac{2}{1-v}dv-\int dv=\int \frac{dx}{x}$

or, $-2\log (1-v)-v=\log +c$

or $2\log (1-v)+\log x+v=c$

or, $\log (1-v{)}^2+\log x+v=c$

or $\log \left\{x\right(1-\frac{y}{x}{)}^2\}+\frac{y}{x}=c$

or $\log \left\{x\right(\frac{x-y}{x}{)}^2\}=c-\frac{y}{x}$

or $(x-y{)}^2.\frac{1}{x}=c.e^{-y/x}$

or $(x-y{)}^2=x{e}^{-y/x}.c$