first order homogeneous differential equation confusion

Question

So I have the differential equation $(y^2+xy)dx+(3xy+x^2)dy=0$. I am asked to solve this ODE by using the substitution $v=\frac{y}{x}$ but I run into some issues if I do that.

If $v=\frac{y}{x}$ then $y'(x)=v+xv'$

I then divided every term by $x^2$ to get $(\frac{y^2}{x^2}+\frac{y}{x})+(\frac{3y}{x}+1)(\frac{dy}{dx})=0$

If I substitute, I then get, $(v^2+v)+(3v+1)(v+xv')=0$

Now I run into the problem where I can't make this into a separable equation...

I know that this is an exact equation and I can solve it if I use an integration factor $u(x)=y$ but I am asked to solve this question using the substitution $v=\frac{y}{x}$. Even if I muliply each term by the integration factor, things become even worse...

Any guidance would be appreciated. Thanks.

Answer 1

\begin{align} (y^2+xy)dx+(3xy+x^2)dy&=0\\ &\frac{dy}{dx}=\frac{-(y^2+xy)}{(3xy+x^2)}\\ &\text{put}y=vx\implies y'=v+xv'\\ &x\frac{dv}{dx}=\frac{-v^2-v}{3v+1}-v=\frac{-2v(2v+1)}{3v+1}\\ &\int\frac{3v+1}{-2v(2v+1)}dv=\int\frac{dx}{x}\\ &\int\frac{(2v+1)+(v)}{-2v(2v+1)}dv=\int\frac{dx}{x}\\ &\int\frac{-1}{2v}dv-\frac{1}{2}\int\frac{1}{2v+1}=\log x+\log c\\ &\frac{-1}{2}\log v-\frac{1}{4}\log ({2v+1})=\log cx\\ &\frac{1}{\sqrt v.(2v+1)^{1/4}}=cx\\ &\text{just put the value of $v=\frac yx$.} \end{align}

Answer 2

In the equation $\underbrace{(y^2+xy)}_{P}dx+\underbrace{(3xy+x^2)}_{Q}dy=0,$ the functions $P$ and $Q$ are homogeneous with the same degree $d$ (grade 2 in this case). By a well-kown property, the substitution $y=vx$ transforms the given equation in other one with separate variables.

Proof. $\; y=vx\Rightarrow P(x,y)=P(x,vx)\;,\;Q(x,y)=Q(x,vx).$
As $P$ and $Q$ are homogeneous with degree $d$: $$P(x,vx)=P(1\cdot x,vx)=x^dP(1,v)=x^dR(v),$$ $$Q(x,vx)=Q(1\cdot x,vx)=x^dQ(1,v)=x^dS(v).$$ We get $\;\left(R(v)+vS(v)\right)dx+xS(v)dv=0\text{ (separate variables)}$