Question

i have equation

Answer 1

Solve the homogenous part of the equation first.
\[y''+4y'+4y=0\]
That'll give you your general solution

Answer 2

\[y''+4y'+4y=0\]
\[Y^2+4Y+4= (Y+2)^2\]
\[y_{c}=c_{1}e^{-2t}+c_{2}te^{-2t}\]
Now that you have your general solution, plug it back into the original equation.
\[y_{c}''+4y_c'+4y_{c}=t^2e^{-2t}\]
You can find all the derivatives separately
\[4y_{c}=4c_{1}e^{-2t}+4c_{2}te^{-2t}\]
\[4y_{c}'=-8c_{1}e^{-2t}-8c_{2}te^{-2t}+4c_{2}e^{-2t}=(-8c_{1}+4c_{2})e^{-2t}-8c_{2}te^{-2t}\]
\[y_{c}''=\frac{d(-2c_{1}e^{-2t}-2c_{2}te^{-2t}+c_{2}e^{-2t})}{dt}=4c_{1}e^{-2t}+4c_{2}te^{-2t}-2c_{2}e^{-2t}-2c_{2}e^{-2t})\]
simplified
\[y_{c}''=(4c_{1}-4c_{2})e^{-2t}+4c_{2}te^{-2t}\]
\[y_{c}^2+4y_{c}+4y_{c}= (4c_{1}-4c_{2})e^{-2t}+4c_{2}te^{-2t} + (-8c_{1}+4c_{2})e^{-2t}-8c_{2}te^{-2t} +4c_{1}e^{-2t}+4c_{2}te^{-2t}=t^2e^{-2t}\]
Combine like terms and solve for the coefficients.