What is the solution for y^{''}-4y^'+4y=t^3e^{2t},y(0)=0,y^'(0)=0 ?

The solution for y^{''}-4y^'+4y=t^3e^{2t},y(0)=0,y^'(0)=0 is y=(e^{2t}t^5)/(20)

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y^{''}-4y^'+4y=t^3e^{2t},y(0)=0,y^'(0)=0

y^{^{''}} - 4 y^{^{'}} + 4 y = t^{3} e^{2 t}, y (0) = 0, y^{^{'}} (0) = 0

y = \frac{e ^{2 t} t ^{5}}{20}

Solution steps

y^{''} (t) - 4 y^{'} (t) + 4 y = t^{3} e^{2 t}

Solve linear ODE:

y = \frac{e ^{2 t} t ^{5}}{20}

y = \frac{e ^{2 t} t ^{5}}{20}

f (x) = cot (x)

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

(x^{2} y + y) y^{^{'}} = 2 x y^{2} + 6 x

\frac{d}{d x} (\frac{( 2 π x ^{2} + π x + \frac{π}{8} )}{24 x})

\frac{\partial}{\partial s} (\frac{r}{s})

What is the solution for y^{''}-4y^'+4y=t^3e^{2t},y(0)=0,y^'(0)=0 ?
The solution for y^{''}-4y^'+4y=t^3e^{2t},y(0)=0,y^'(0)=0 is y=(e^{2t}t^5)/(20)