Question

Show that the differential equation $(x^2+xy)dy=(x^2+y^2)dx$ is homogeneous and solve it.

Answer 1

Given differential equation :

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

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

If $F(\lambda x,\lambda y)=\lambda F(x,y),$ then the
differential equation is homogeneous.

Substituting $F(x,y)=\frac{x^2+y^2}{x^2+xy}$
Now,

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

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

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

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

Hence, the differential equation is homogeneous.

Given differential equation is

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

$\frac{dy}{dx}=\frac{x^2{y}^2}{x^2+xy}...\left(i\right)$

$Lety=vx$

differentiate both sides $w.r.t{}^{\prime }{x}^{\prime },$ we get

$\frac{dy}{dx}=\frac{d\left(vx\right)}{dx}$

Using product rule :$(uv{)}^{\prime }=u^{\prime }v+u{v}^{\prime }$

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

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

$\text{Substituting the value of}\frac{dy}{dx}andy=vxin\left(i\right),weget$

$x\frac{dv}{dx}+v=\frac{x^2+(vx{)}^2}{x^2+x\left(vx\right)}$

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

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

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

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

$\frac{(1+v)}{(v-1)}dv=\frac{-dx}{x}$

Integrating both sides, we get

$\int \frac{(1+v)}{(v-1)}dv=\int \frac{-dx}{x}$

$\int \frac{(1+v)}{(v-1)}dv=-\log |x|+c$

$I=-\log |x|+c...\left(ii\right)$

$I=\int \frac{(1+v)}{(v-1)}dv$

$I=\int \left(\frac{v-1+2}{v-1}\right)dv$

$I=\int (\frac{v-1}{v-1}+\frac{2}{v-1})dv$

$I=\int dv+\int \frac{2}{v-1}dv$

$I=v+2\log |v-1|$

$\text{Substituting}v=\frac{y}{x},weget$

$I=\frac{y}{x}+2\log |\frac{y}{x}-1|$

$I=\frac{y}{x}+2\log \left|\frac{y-x}{x}\right|$

Substituting the value of $I$ in (ii), we get

$\frac{y}{x}+2\log \left|\frac{y-x}{x}\right|=-\log |x|+c$

$\frac{y}{x}+\log \left|\frac{(y-x{)}^2}{x^2}\right|+\log |x|=c$

$(\because n\log b=\log b^n)$

$\frac{y}{x}+\log |\frac{(y-x{)}^2}{x^2}\times x|=c$

$(\because \log a+\log b=\log ab)$

$\log \left|\frac{(y-x{)}^2}{x}\right|=c-\frac{y}{x}$

$\frac{(y-x{)}^2}{x}=e^c\frac{y}{x}$

$\frac{(y-x{)}^2}{x}=e^c\times e^{\frac{y}{x}}$

Substituting $e^c=k$

$\frac{(y-x{)}^2}{x}=k{e}^{\frac{y}{x}}$

$(x-y{)}^2=kx{e}^{\frac{y}{x}}$

Hence, the required general solution is

$(x-y^2)=kx{e}^{\frac{y}{x}}$