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Given the ode: $$y''-2y'+y=e^t$$

How can I find the form of the particular solution?

I tried $y=Ae^t$ but

$frac{d^2y}{dt^2}Ae^t=frac{dy}{dt}Ae^t=Ae^t$

$Ae^t-2Ae^t+Ae^t=e^t$

$0=e^t$

So this doesn't work

I also tried $y=Ate^t$ but again

$frac{d^2y}{dt^2}=A(2e^t+te^t)$

$frac{dy}{dt}=A(e^t+te^t)$

$A(2e^t+te^t)-2A(e^t+te^t)+Ate^t=e^t$

$2A+At-2A-2At+At=1$

again doesn't work

GEnerally what is the best way to guess the form of the solution?

Hints/Guides on how to solve such differential equations :

$mathbf{1}$ - Method of Undetermined Coefficients :

Start of by solving the homogenous equation $y''-2y'+y= 0$ by assuming that a solution will be proportional to $e^{lambda t}$ for some $lambda$. Substitute in and calculate $lambda$. Notice the multiplicity of the solution for $lambda$ and adjust your general solution accordingly.

Then, use the method of undetermined coefficients to find a particular solution of the problem for $y''-2y'+y=e^t$.

The general solution of the initial differential equation, will then be the general solution of the homogenous plus the particular solution you found.

You can find more information and examples about that method, here.

$mathbf{2}$ - Laplace Transformation :

This is a very fast and straight forward way to approach the problem, but it needs some fluent handling of Laplace Transformation techniques. Note that you can apply the Laplace Transformation without even needing initial conditions, simply stating them as constants.

Start of by applying the Laplace Transformation

$$mathcal{L}_tbig[f(t)big](s) = int_0^infty f(t)e^{-st}mathrm{d}t$$

to both sides of the given differential equation :

$$mathcal{L}_tbig[y'' - 2y' + y'] = mathcal{L}_t[e^t]$$

$$Leftrightarrow$$

$$(s-1)^2big[mathcal{L}_t[y(t)](s)big] - (s-2)y(0) - y'(0) = frac{1}{s-1}$$

$$Leftrightarrow$$

$$mathcal{L}_tbig[y(t)big](s) = frac{y(0)(s^2-3s+2) + y'(0)(s-1) + 1}{(s-1)^3}$$

$$=$$

$$mathcal{L}_tbig[y(t)big](s) = frac{1}{(s-1)^3} - frac{y(0)}{(s-1)^2} + frac{y(0)}{s-1} + frac{y'(0)}{(s-1)^2} $$

$$implies$$

$$y(t) = frac{1}{2}e^t(t^2+2c_1 - 2c_1t + 2c_2t) = frac{e^tt^2}{2} + c_1e^t - c_1e^tt + c_2e^tt$$

$mathbf{3}$ - Variation of Parameters :

You must repeat the step of solving the homogenous equation by finding that $lambda$s mentioned. Then by listing the basis solution as $y_{b_1} = e^t$ and $_{b_2} = e^tt$ you can use variation of parameters to find the final general solution by computing the Wronskian and finding the integrals :

$$v_1(t) = - int frac{f(t)y_{b_2}(t)}{W(t)}mathrm{d}t quad text{and} quad v_2(t) = int frac{f(t)y_{b_1}(t)}{W(t)}mathrm{d}t$$

You can find more information and examples about that method, here.

Both your attempts are in fact right but fail because the fundamental set of solutions for your second order ODE is given by exactly your both guesses for the particular solution. It is not hard to show by using the characteristic equation that the fundamental set of solutions is given by

$$y(t)=c_1e^t+c_2te^t$$

Therefore in this case it is not possible to obtain a particular solution in the standard way hence the inhomogeneous term is in fact part of the solution.

Howsoever you second try was near the actual solution. Since both terms $e^t$ and $te^t$ did not worked out hence there are part of the solution you could further consider $t^2e^t$ as the next try. It turns out that this yields to the solution $($see here$)$.

This leads to the conjecture that at least for a inhomogeneous term of the form $Ae^{bt}$ you just have to try the specific exponential with different powers of the variable infront as particular solution until it works out.