I got the sum of A is 0? There is

I got the sum of A is 0? There is no solution to this? Can someone please help. Tha
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$$\displaystyle{y}{''}-{4}{y}'+{4}{y}=-{6}{e}^{{{2}{t}}}$$

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Foreckije
Hint: make for the particular solution the ansatz
$$\displaystyle{y}_{{p}}={A}{e}^{{{2}{t}}}{t}^{{2}}$$
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porschomcl
I have no idea what you mean by "sum of A". There is no "A" in the problem. Do you mean that you are looking for a "particular solution" of the form $$\displaystyle{A}{e}^{{{2}{t}}}$$? If so you should NOT be getting A= 0- you should not be getting anything at all for A! Setting $$\displaystyle{y}={A}{e}^{{{2}{t}}},{y}'={2}{A}{e}^{{{2}{t}}}$$ and $$\displaystyle{y}{''}={4}{A}{e}^{{{2}{t}}}$$ so that $$\displaystyle{y}{''}-{4}{y}+{4}={4}{a}{e}^{{{2}{t}}}-{8}{A}{e}^{{{2}{t}}}+{4}{A}{e}^{{{2}{t}}}={0}$$ no matter what A is
The "double root" that Bruce refers to is the double root of the characteristic equation: the associated homogeneous differential equation, $$\displaystyle{y}{''}-{4}{y}'+{4}{y}={0}$$, has characteristic equation $$\displaystyle{r}^{{2}}-{4}{y}+{4}={\left({r}-{2}\right)}^{{2}}={0}$$ which has the double root r= 2. The general solution to the associated homogeneous equation is $$\displaystyle{y}{\left({t}\right)}={C}_{{1}}{e}^{{{2}{t}}}+{C}_{{2}}{t}{e}^{{{2}{t}}}$$. You need to look for a specific solution of the form $$\displaystyle{y}{\left({t}\right)}={A}{t}^{{2}}{e}^{{{2}{t}}}$$