Question

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

• A. $\log x=\log (x-y)+\frac{y}{x}+c$
• B. $\log x=2\log (x-y)+\frac{y}{x}+c$
• C. $\log x=\log (x-y)+\frac{x}{y}+c$
• D. $\log x=2\log (x-y)+\frac{x}{y}+c$

Answer 1

The correct option is B $\log x=2\log (x-y)+\frac{y}{x}+c$

$(x^2+xy)dy=(x^2+y^2)dx$ or $\frac{dy}{dx}=\frac{x^2+y^2}{x^2+xy}$
Let $\frac{y}{x}=v$. Then $\frac{dy}{dx}=v+x\frac{dv}{dx}$
Thus, equation reduces to
$x\frac{dv}{dx}=\frac{1+v^2}{1+v}-v$
$=\frac{1+v^2-v-v^2}{1+v}$
$=\frac{1-v}{1+v}$
or $\int \frac{1+v}{1-v}dv=\int \frac{dx}{x}$
or $-\int (1-\frac{2}{1-v})dv=\int \frac{dx}{x}$
or $-v-2\log (1-v)=\log x+\log c$
or $-\frac{y}{x}-2\log \left(\frac{x-y}{x}\right)=\log x+\log c$
or $\frac{-y}{x}-2\log (x-y)+2\log x=\log x+\log c$
or $\log x=2\log (x-y)+\frac{y}{x}+k$ where $k=\log c$