Solving a homogenous differential equation

Question

I am taking a course on differential equations. In my study booklet, I am given the following problem:

Solve the differential equation $$\frac{dy}{dx}=\frac{2x+y}{3x+2y}$$

My solution.

$(2x + y)dx - (3x+2y)dy = 0$

This differential equation is not separable. Substituting $y=ux$, $dy=u dx + x du$, we obtain :

$P(x,ux)=2x+ux=x(2+u)$

$Q(x,ux)=3x+2ux=x(3+2u)$

If $x\neq 0$, we have :

$\begin{align} (2 + u)dx &- (3 + 2u)dy &= 0 \\ (2 + u)dx &- (3 + 2u)(udx+ xdu) &= 0\\ (2 + u)dx &- (3u + 2u^2)dx - (3 + 2u)xdu &=0\\ (2 - 2u - 2u^2)dx &- (3+2u)xdu &= 0\\ \frac{dx}{x} &- \frac{3+2u}{2 - 2u - 2u^2}du &= 0 \end{align}$

This differential equation is now in the separable form. It's solution is :

$ \begin{align} \int\frac{dx}{x} &- \int\frac{3+2u}{2 - 2u - 2u^2}du &= c \end{align} $

Further,

$ \begin{align} (2 - 2u - 2u^2)'&= 2(1 - u - u^2)' \\ &= 2(-1 - 2u) \\ &= -2 - 4u \\ \end{align} $

$\boxed {\therefore (3 + 2u) = -\frac{1}{2}(-2 - 4u) + 2}$

And,

$ \begin{align} (2 - 2u - 2u^2) &= -2(u^2 + u - 1) \\ &= -2\{(u+1/2)^2-5/4\} \\ &= -2\{(u+1/2)^2-(\sqrt 5/2)^2\} \\ &= 2\{(\sqrt 5/2)^2-(u+1/2)^2\} \\ \end{align} $

$\boxed {(2 - 2u - 2u^2) = 2\{(\sqrt 5/2)^2-(u+1/2)^2\}}$

Making these substitutions into the integral, we obtain :

$ \begin{align} \int\frac{dx}{x} &- \int\frac{(-\frac{1}{2})(-2-4u)}{2 - 2u - 2u^2}du &- \int\frac{2}{2\{(\sqrt 5/2)^2-(u+1/2)^2\}}du &= c \\ \log x &+ \frac{1}{2} \log (2 - 2u - 2u^2) &- \frac{1}{2(\sqrt {5}/2)}\log \left({\frac{u+\frac{1}{2}-\frac{\sqrt 5}{2}}{u+\frac{1}{2}+\frac{\sqrt 5}{2}}}\right) &= c \\ & \log (2x^2 - 2xy - 2y^2) & - \frac{2}{\sqrt 5} \log \left(\frac{2y + x(1 - \sqrt 5)}{2y + x(1 + \sqrt 5)}\right) &= c \end{align} $

However, my solution doesn't match with what's given in the book. Could someone please verify my steps?

Answer 1

Hint: Substituting $$y(x)=v(x)x$$ then we get $$\frac{\frac{dv(x)}{dx}(v(x)+3)}{-v(x)^2-2v(x)+2}=\frac{1}{x}$$ and integrate. The solution should be $$\text{Solve}\left[\frac{1}{5} \left(5+2 \sqrt{5}\right) \log \left(-\frac{2 y(x)}{x}+\sqrt{5}-1\right)+\frac{1}{5} \left(5-2 \sqrt{5}\right) \log \left(\frac{2 y(x)}{x}+\sqrt{5}+1\right)=c_1-2 \log (x),y(x)\right]$$

Answer 2

The solution and method given by Dr. SG should be correct which has one constant of integration $c$ and rightly so, you get the general solution of your ODE. But, I want to bring your attention to the fact that in your ODE if you put $y=vx$ but treat $v$ as contant, you get two roost of $v$ as $\frac{-1\pm\sqrt5}{2}$. So your ODE has two singular (fixed or essential) solutions without any constant of integration which are $y=v_1x$ and $y=v_2x$. These singular solutions are often not looked for. Can a first order homogeneous ODE have singular solution(s) (free of a constant)