I've got differential equation.
$x^2 \cdot \frac{dy}{dx} + xy + 1 = 0$
I know it's linear differential equation. Putting this equation into calculator, I've seen interesting substitution solution:
- $y = z^{\lambda}, x = z$, which, in turn, leads us to equation
- $\lambda z^{\lambda + 1} + z^{\lambda + 1} + 1 = 0$
Then calculator makes a comparison $\lambda + 1 = \lambda + 1 = 0 \implies \lambda = -1$
Another equation:
$2 \cdot \frac{dy}{dx} + x = 4 \sqrt{y}$
is began solution in the same way.
After substitution, it gets
$2 \lambda z^{\lambda - 1} + z = 4 z^{\frac{\lambda}{2}}$
and compares in the same way:
$\frac{\lambda}{2} = \lambda - 1 = 1 \implies \lambda = 2$
The most interesting case, in my opinion, is trying to solve:
$2xy \cdot \frac{dy}{dx} = 3 \sqrt{x^6 - y^4} + 3y^2$, which turns into
$2 \lambda z^{2 \lambda} = 3\sqrt{z^6 - z^{4 \lambda}} + 3z^{2 \lambda}$, but now comparison is a system(?):
$\left\{\begin{matrix} 4 \lambda = 6\\ 3 = 2 \lambda = 2 \lambda \end{matrix}\right. \implies \lambda = \frac{3}{2}$
In all three cases calculator has got right answers after another additional substitution $z = ux$
So my question is.... Why does it compare powers in such way? Where did we get this equations from? Because, for example, if we put $\lambda = -1$ in first equation, we get
2) $\lambda z^{-1 + 1} + z^{-1 + 1} + 1 = \lambda + 2 = 0 \implies \lambda = -2$. A little bit controversial. So what's going on?