Question

Solve the following differential equation: $2xydx+\left(x^2+2y^2\right)dy=0.$

Answer 1

Solve the following differential equation.

$(x^2-y^2)dy+2xy.dx=0$.

Sol:- To find $dy/dx$

$(x^2-y^2)dx+2xy.dy=0$

$2xy.dy=-(x^2-y^2).dx$

$2xy.dy=(y^2-x^2).dx$

$dy/dx=\frac{y^2-x^2}{2xy}$ …………..$\left(1\right)$

Putting $F(x,y)=dy/dx$ and finding $F=(\lambda x,\lambda y)$

$F(x,y)=\frac{y^2-x^2}{2xy}$

$F(\lambda x,\lambda y)=\frac{(\lambda y{)}^2-(\lambda x{)}^2}{2xy\lambda }=\frac{{\lambda }^2.y^2-{\lambda }^2.x^2}{{\lambda }^2.2xy}=\frac{{\lambda }^2(y^2-x^2)}{{\lambda }^2.2xy}$

$F(\lambda x,\lambda y)=\frac{y^2-x^2}{2xy}=F(x,y)$

$\therefore F(\lambda x,\lambda y)=F(x,y)={\lambda }^o.F(x.y)$

Hence, $F(x,y)$ is a homogenous function of with degree zero.

So, $dy/dx$ is a differentiate equation solving $dy/dx$ by putting $y=vx$

Put $y=vx$

Differentiating w.r.t. x

$dy/dx=x.dv/dx+v.dx/dx$

$dy/dx=x.dv/dx+v$

Putting value of $dy/dx$ and $y=vx$ in $\left(1\right)$

$\frac{dy}{dx}=\frac{y^2-x^2}{2xy}$

$x.dv/dx+v=(xv{)}^2-x^2/2x(xv)$

$\Rightarrow x.dv/dx+v=x^2.v^2-x^2/2x^2.v$

$\Rightarrow x.dv/dx=x^2.v^{-}{x}^2/2x^2.v$

$x.dv/dx=x^2.v^2-x^2-2.v^2.x^2/2x^2.v$

$x.dv/dx=-x^2.v^2-x^2/2x^2.v$

$dv/dx=\frac{1}{x}\left(\frac{x^2(v^2-1)}{2x^{2}v}\right)$

$dv/dx=1/x.(v^2+1/2v)$

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

Integrating both sides

$\int 2v/v^2+1.dv=\int -dx/x$

$\int 2v/v^2+1.dv=-log|x|+c$ ……………..$\left(2\right)$

Putting $t=v^2+1$, differentiating w.r.t. x

$d/dv(v^2+1)=dt/dv$

$2v=dt/dv$

$dv=dt/2v$

Now from $\left(2\right)$

$\int 2v/t^2.dt/dv=-log|x|+c$

$\int dt/t=-log|x|+c$

$log|t|=-log|x|+c$

Putting $t=v^2+1$

$log|v^2+1|=-log|x|+c$

$log|v^2+1|+log|x|=c$

$log|x(v^2+1)|=c$

Putting $v=y/x$

$log|\left[x\right((y/x{)}^2+1)]|=c$

Putting $c=logc$

$log|y^2+x^2||x|-logc$

$y^2+x^2=cx$

$x^2+y^2=cx$

Is the general solution of differential equation.