# Second-order linear homogenous ODE - I can't see any substitution!

$$y'' - y' + e^{2x}y=0$$

At first it looks like it's an easy ODE, but I haven't been able to solve it.

For example, if I let $$y=u(x)\cdot z, z(x)$$ and accordingly choose $$u$$ so that the coefficient of $$z'$$ (after performing the necessary derivations and grouping all terms by the order of derivation of $$z$$) is zero, I get the following equation:

$$z'' + (e^{2x} - \frac{1}{2})z = 0$$ which isn't any easier to solve than the first one.

I'm having trouble with this equation because I don't think there's an apparent substitution to be done. Can anyone help? Thanks!

$$y'' - y' + e^{2x}y=0$$ Note that : $$y'=\dfrac {dy}{dx}=\dfrac{dy}{de^{x}}\dfrac {de^x}{dx}=e^{x}\dfrac {dy}{de^{x}}=u\dfrac {dy}{du}$$ Find $$y''$$ then simplify the DE and solve. $$\dfrac {d^2y}{du^2}+y=0$$ where $$u=e^x$$. The final answer should be: $$y(x)=c_1 \cos u+c_2 \sin u$$ $$y(x)=c_1 \cos (e^x)+c_2 \sin (e^x)$$

Hint.

We can find a particular solution with the structure $$y_p = e^{\alpha e^t}$$ because

$$y''_p - y'_p+e^{2t}y_p = (\alpha^2+1)e^{\alpha e^t+2t}$$