$y''-y'+e^{2x}y=0$
can you give me a hint which I can start with
Should I use the fact that if $r_1$ and $r_2$ are complex numbers (which happens if $b^2 – 4ac < 0$), then the general solution is
$$y = c_1 \Bbb e ^{\alpha x} \cos \beta x + c_2 \Bbb e ^{\alpha x} \sin \beta x$$
where
$$r_1 = \bar r_2 = \alpha + \Bbb i \beta = \frac {-b} {2a} + \Bbb i \frac {\sqrt {4ac – b^2}} {2a} ,$$
or make $t=e^x$?
Best Answer
There are often multiple ways to get to the final solution of a differential equation. I used $t=ie^x$ and got to $y = c_1 \cos(t) + c_2 \sin t$, yielding the solution $y = c_1 \cosh(e^x) + c_2 \sinh(e^x)$. I haven't checked it, but the substitution $t = e^x$ might work similarly. As a tip, if you show something you have attempted you could likely receive more help, as I (and likely others) don't feel like TeXing a proof for solving that Dif-Eq. Try it your way and let us know where you get stuck.
First few steps I did, using $t = ie^x$ and $\frac{dt}{dx} = t$. First is the original equation with substitution $$\frac{d^2 y}{dx^2} - \frac{dy}{dx} = -t^2y$$ Apply chain rule for both derivatives $$\frac{d^2 y}{dx^2} = \frac{d}{dx} \bigg(\frac{dy}{dt}\frac{dt}{dx}\bigg) = \bigg(\frac{dt}{dx}\bigg)^2\frac{d^2 y}{dt^2} + \frac{d^2t}{dx^2}\frac{dy}{dt} = t^2\frac{d^2 y}{dt^2} + t \frac{dy}{dt}$$ $$\frac{dy}{dx} = \frac{dt}{dx}\frac{dy}{dt} = t \frac{dy}{dt}$$ Substituting back in, we get $$\Rightarrow t^2\frac{d^2 y}{dt^2} + t \frac{dy}{dt} - t \frac{dy}{dt} = -t^2y$$ $$\Rightarrow t^2\frac{d^2 y}{dt^2} = -t^2y$$ $$\Rightarrow \frac{d^2 y}{dt^2} = -y$$
Can you get it from here?