Q29: Solve $y\,dx + (x - y^2)\,dy = 0$⁠, or $x\,dy - y\,dx = \sqrt{x^2 + y^2}\,dx$

Because of the $y^2$ term, the equation is not linear in $y$⁠. But $x$ appears only to the first power, so it is linear in $x$⁠. So we treat $x$ as the dependent variable and $y$ as the independent one.

The linear form is then $\frac{dx}{dy} + Px = Q$⁠, where $P$ and $Q$ are functions of $y$ only.

Then add $\frac{x}{y}$ to both sides to get $\frac{dx}{dy} + \frac{x}{y} = y$⁠.

The integrating factor is $e^{\int P\,dy}$⁠. Here $\int\frac1y\,dy = \log y$ (taking $y > 0$⁠), and $e^{\log y} = y$ because $e^x$ and $\log x$ undo each other.

For $\frac{dx}{dy} + Px = Q$ the solution is

Here $Q \cdot \text{I.F.} = y \cdot y = y^2$⁠, and $\int y^2\,dy = \frac{y^3}{3}$⁠. Watch out: add the constant $C$ at this step, not later.

A differential equation is homogeneous when $\frac{dy}{dx}$ can be written as a function of $\frac{y}{x}$ alone. Divide the numerator by $x$⁠; for $x > 0$⁠, $x = \sqrt{x^2}$⁠, so

Step by step:

For a homogeneous equation the substitution is $y = vx$⁠. By the product rule, $\frac{d}{dx}(vx) = v + x\frac{dv}{dx}$⁠.

The $v$ on both sides cancels. Then put everything with $v$ on one side and everything with $x$ on the other: divide by $x\sqrt{1 + v^2}$ and multiply by $dx$⁠.

On the right, $\frac{y}{x} = v$⁠, so $\sqrt{1 + \left(\frac{y}{x}\right)^2} = \sqrt{1 + v^2}$⁠.

Standard integrals (the first with $a = 1$⁠):

Writing the constant as $\log K$ lets us combine the logs in the next step.

With $x > 0$⁠, $\log|x| + \log K = \log(Kx)$⁠, and

because

The log of a quotient is the difference of the logs, so the left side is $\log\left|y + \sqrt{x^2 + y^2}\right| - \log x$⁠. Add $\log x$ to both sides: $\log(Kx) + \log x = \log(Kx^2)$⁠. Equal logs give equal moduli.

Since $K$ is any positive constant, $\pm K$ is just an arbitrary constant, written $C$⁠.