What is the solution for y^{''}-4y^'+4y=t^3y^0 ?

The solution for y^{''}-4y^'+4y=t^3y^0 is y=c_{1}e^{2t}+c_{2}te^{2t}+(t^3)/4+(3t^2)/4+(9t)/8+3/4

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y^{''}-4y^'+4y=t^3y^0

y^{^{''}} - 4 y^{^{'}} + 4 y = t^{3} y^{0}

y = c_{1} e^{2 t} + c_{2} t e^{2 t} + \frac{t ^{3}}{4} + \frac{3 t ^{2}}{4} + \frac{9 t}{8} + \frac{3}{4}

Solution steps

y^{''} (t) - 4 y^{'} (t) + 4 y = t^{3} y^{0}

Solve linear ODE:

y = c_{1} e^{2 t} + c_{2} t e^{2 t} + \frac{t ^{3}}{4} + \frac{3 t ^{2}}{4} + \frac{9 t}{8} + \frac{3}{4}

y = c_{1} e^{2 t} + c_{2} t e^{2 t} + \frac{t ^{3}}{4} + \frac{3 t ^{2}}{4} + \frac{9 t}{8} + \frac{3}{4}

\frac{d ^{2}}{d x ^{2}} (e^{3 x} cos (2 x))

\frac{d}{d x} (9 sin (x) cos (x))

\frac{1}{s ^{2} + 2 ^{2}}

\frac{d x}{d t} = - \frac{5 x}{100}

\int sin (2 x) - sin (\frac{x}{2}) d x

What is the solution for y^{''}-4y^'+4y=t^3y^0 ?
The solution for y^{''}-4y^'+4y=t^3y^0 is y=c_{1}e^{2t}+c_{2}te^{2t}+(t^3)/4+(3t^2)/4+(9t)/8+3/4