Suppose we have a very standard form of differential equation
$$Mdx+Ndy=0$$
Which is homogeneous.
Show that the general solution of the above equation can be written as below,
$$x=c \phi (\frac{y}{x})$$ ,c is an arbitrary constant!
My attempt is as below:
We know that a homogeneous equation can be transformed into a separable equation with substitution of $y=vx$!$\frac{dy}{dx}=v+x\frac{dv}{dx}$
Knowing that $$N(tx,ty)=t^nN(x,y)$$ $$M(tx,ty)=t^nM(x,y)$$
Setting $t=\frac{1}{x}$
We would have $$\frac{dy}{dx}=-\frac{M(1,\frac{y}{x})}{N(1,\frac{y}{x})}$$
In standard form we have${\frac{dy}{dx}=f(\frac{y}{x})}$
$$v+x\frac{dv}{dx}=f(v)$$
Integrating we have
Let $$\int \frac{dv}{f(v)-v}=ln|\phi v|$$
$$x=c(\phi v)$$
$$x=c(\phi \frac{y}{x})$$
Can someone verify my work. If it is not enough please add it in the answer.
Your work is correct. Here is how I would do it: We have a homogeneous differential equation. This means, we can write
$$M(x,y) = \sum_{m=1}^n a_mx^my^{n-m} = x^n\sum_{m=1}^n a_mx^{m-n}y^{n-m} = x^n\sum_{m=1}^n a_m\left(\frac{y}{x}\right)^{n-m} $$
For $N(x,y)$, we can do exactly the same thing.
Now:
$$Mdx + Ndy = 0 \iff Mdx = -Ndy \iff \frac{M}{N} = \frac{-dy}{dx}$$ $$\Rightarrow \frac{x^n\sum_{m=1}^n a_m\left(\frac{y}{x}\right)^{n-m}}{x^n\sum_{m=1}^n b_m\left(\frac{y}{x}\right)^{n-m}}dx = -dy$$
$$\Rightarrow \phi\left(\frac{y}{x}\right)dx = -dy$$
Now make the substitution $$z = \frac{y}{x}, zx = y, dy = dzx + zdx$$
so the differential equation becomes:
$$\phi(z)dx = -(xdz + zdx)$$
or equivalently:
$$(\phi(z) + z)dx = -xdz$$
$$\Rightarrow \int\frac{dx}{x} = \int\frac{-dz}{\phi(z)+z}$$
$$\Rightarrow \ln|x| = \int\frac{-dz}{\phi(z)+z}$$
$$\Rightarrow x = c \exp\left(\int\frac{-dz}{\phi(z)+z}\right) = c\psi(z) = \boxed{c\psi\left(\frac{y}{x}\right)}$$