Question

For each of the differential equations in Exercises from $11$ to $15$, find the particular solutions satisfying the given condition:
$(x+y)dy+(x-y)dx=0;y=1$ when $x=1$

Answer 1

$(x+y)dy+(x-y)dx=0$

$\frac{dy}{dx}=\frac{-(x-y)}{x+y}$

Let $f(x,y)=\frac{dy}{dx}=\frac{-(x-y)}{x+y}$

$f(\lambda x,\lambda y)=\frac{-(\lambda x-\lambda y)}{\lambda x+\lambda y}-\frac{-(x-y)}{x+y}$

$={\lambda }^{o}f(x,y)$

$\therefore f(x,y)$ is a homogeneous function

Put $y=vx$

Diff w.r.t. ${}^{\prime }{x}^{\prime }$

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

$\frac{dy}{dx}=-(x-y)/x+y$

$v+\frac{xdv}{dx}=\frac{-(x-vx)}{x+vx}$

$v+\frac{xdv}{dx}=\frac{v-1}{1+v}$

$\frac{xdv}{dx}=\frac{v-1}{1+v}-v=\frac{v-1-v-v^2}{1+v}$

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

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

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

$\int \frac{1}{v^2+1}.dx+\int \frac{v}{1+v^2}=-\log x+c$.

$\int \frac{v}{v^2+1}dv+{\tan }^{-1}v=-\log x+c$.

$v^2+1=t.$

$2vdv=dt$

$vdv=dt/2$

${\tan }^{-1}v+\frac{1}{2}\log t=-\log x+c.$

${\tan }^{-1}v+\frac{1}{2}\log (v^2+1)=-\log x+c.$

${\tan }^{-1}\frac{y}{x}+\frac{1}{2}\log (\frac{y^2}{x^2}+1)=-\log x+c.$

${\tan }^{-1}\frac{y}{x}+\log \sqrt{\frac{x^2+y^2}{x^2}}+\log x=c$

${\tan }^{-1}\frac{y}{x}+\log \frac{\sqrt{x^2+y^2}}{x}+\log x=c.$

${\tan }^{-1}\frac{y}{x}+\log \sqrt{x^2+y^2}=c$

Put $x=1$ & $y=1$

${\tan }^{-1}\left(\frac{1}{1}\right)+\log \sqrt{1+1}=c$.

${\tan }^{-1}1+\log \sqrt{2}=c$.

$\frac{\pi }{4}+\frac{1}{2}\log 2=c$.

Put value of $C$ in $\left(1\right)$

${\tan }^{-1}\frac{y}{x}+\log \sqrt{x^2+y^2}=\frac{\pi }{2}\log 2$

${\tan }^{-1}\frac{y}{x}+\frac{1}{2}\log (x^2+y^2)=\frac{\pi }{4}+\frac{1}{2}\log 2$

Multiplying by $2$ on $B.S.$

$\log (x^2+y^2)+2{\tan }^{-1}\frac{y}{x}=\frac{\pi }{2}+\log 2.$