Browse Questions

# Solve the differential equation $(x^2-yx^2)dy+(y^2+x^2y^2)dx=0$ Given $y=1$ when $x=1$

$(a)\;(x^{2}+x-1)y=(1+ylogy)x\qquad(b)\;(x^{2}+x+1)y=(1+ylogy)x\qquad(c)\;(x^{2}+x-1)y=(1+ylogy)x\qquad(d)\;(x^{2}+x+1)y=(1-ylogy)x$

Given equation is $(x^2-yx^2)dy+(y^2+x^2y^2)dx=0$
$\Rightarrow\:\large\frac{dy}{dx}=-\large\frac{y^2+x^2y^2}{x^2-yx^2}$
Taking $y^2$ common from numerator and $x^2$ common from denominator we get
$\large\frac{dy}{dx}=-\large\frac{y^2(1+x^2)}{x^2(1-y)}$
$\Rightarrow\:\large\frac{1-y}{y^2}$$dy=-\large\frac{1+x^2}{x^2}$$dx$
Integrating both the sides
$\Rightarrow\:\int \large\frac{1-y}{y^2}$$dy=-\int\large\frac{1+x^2}{x^2}$$dx$
$\Rightarrow\:\int\large\frac{1}{y^2}.$$dy-\int\large\frac{y}{y^2}$$dy=-\int\large\frac{1}{x^2}$$dx-\int\large\frac{x^2}{x^2}$$dx$
$\Rightarrow\:-\large\frac{1}{y}$$-log y=\large\frac{1}{x}$$-x+c$
Given that $x=1\:\:and\;\:y=1$
$\Rightarrow\:-1-0=1-1+c$
$\Rightarrow\:c=-1$
Hence the solution is $-\large\frac{1}{y}$$-log y=\large\frac{1}{x}$$-x-1$
$\Rightarrow\:(x^2+x-1)y=(1+ylogy)x$