Question

Find the particular solution of the difference equation
$(1-y^2)(1+\log x)dx+2xydy=0$
given that $y=0$ when $x=1$

Answer 1

Consider the following equation.

$(1-y^2)(1+logx)dx+2xydy=0$ and $y=0\&x=1$

$\frac{(1+logx)dx}{x}=\frac{ydy}{(1-y^2)}$

On Integrate both sides, we get

${\int }^{}\frac{(1+logx)dx}{x}={\int }^{}\frac{-2y}{(1-y^2)}dy....\left(1\right)$

Let $u=\log xv=(1-y^2)$

$xdu=dxdy=\frac{1}{-2y}dv$

${\int }^{}\frac{(1+u)xdu}{x}={\int }_i\frac{-2y}{-2yv}dv$

${\int }^{}(1+u)du={\int }^{}\frac{1}{v}dv$

$u+\frac{u^2}{2}=\ln v+C$

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

${\left(\log 0\right)}^2+2\log 0+{\left(1\right)}^2-1=C$

${\left(\log x\right)}^2+2\log x+y^2-1=0$

Hence this is the required answer.