How can I solve $y''-y'+e^{2x}y=0$?

$y''-y'+e^{2x}y=0$ 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 $t=e^x$?

• The method in the gray box is only applicable when the coefficients of the y function and its derivatives are constants. So, that method is not applicable here. – Sinister Cutlass Oct 6 '15 at 5:43

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?

• thanks. can you write what is $\frac{d^2y}{d^2x}$ if $e^x=t$ i mean $\dfrac {d} {dx}\left( \dfrac {tdy} {dt}\right)$ – emmett Oct 6 '15 at 5:52
• @emmett I'm not sure what you are trying to say... I'll write out the first couple steps I used with the substitution $t=ie^x$ and hopefully that will clear some things up for you as you do your substitution. – Brevan Ellefsen Oct 6 '15 at 5:55
• @emmett I have expanded my answer to show how to expand when $t=ie^x$. Does this help you answer your question? It is very similar to what I show, but I leave the work up to you ;) – Brevan Ellefsen Oct 6 '15 at 6:08
• can you please explain this $\dfrac {d} {dx}\left( \dfrac {dy} {dt}\right) =\dfrac {dt} {dx}\left( \dfrac {d^{2}y} {dt^{2}}\right)$ – emmett Oct 6 '15 at 6:13
• I've got to go now, but $$\frac{dt}{dx}\frac{d^2y}{dt^2} = \frac{dt}{dx} \frac{d}{dt} \frac{dy}{dt}= \frac{d}{dt} \bigg(\frac{dt}{dx}\bigg)\frac{dy}{dt} = \frac{d}{dx}\frac{dy}{dt}$$. I haven't checked this, but I think its correct. I've just gotten used to these identities. – Brevan Ellefsen Oct 6 '15 at 6:25

Ansatz : Guess $y = e^{Ax}cos(e^{Bx})$. Then

$y = e^{Ax}cos(e^{Bx})$

$y' = Ae^{Ax}cos(e^{Bx}) - Be^{Ax+Bx}sin(e^{Bx})$.

$y'' =A^2 e^{A x} cos(e^{B x})-B^2 e^{A x+2 B x} cos(e^{B x})-B (2 A+B) e^{A x+B x} sin(e^{B x})$

From the cosine terms, we see that $e^{A+2} + (A+A^2)e^A - B^2e^{A+2B} = 0$.

From the sine terms, we get $-B-2AB - B^2 = 0$.

A solution that will work for the first equation (got from matching the powers) is $A = -1, B=1$. This also satisfies the second equation.

So, a solution is $y = e^{-x}cos(e^x)$. Use reduction of order to get the second linearly independent solution. (If you're tired, you can check it's just $e^{-x}sin(e^x)$.

Let me try with a change of variable, setting $$y=z(t(x))$$ Differentiating, we get $$y'=t'(x)\, z'(t(x))$$ $$y''=t''(x)\, z'(t(x))+t'(x)^2 \,z''(t(x))$$ After simplifications, the differential equation becomes $$t'(x)^2\, z''(t(x))+z'(t(x)) \left(t''(x)-t'(x)\right)+e^{2 x} \,z(t(x))=0$$ It is clear that the equation simplifies if we set $t''(x)=t'(x)$ and/or $t'(x)^2=e^{2x}$ that is to say $t(x)=e^x$. Using it, the equation write $$z''(t)+z(t)=0$$

I am sure that you can take from here.