What is the solution for y^{''}-4y^'+4y=8 ?

The solution for y^{''}-4y^'+4y=8 is y=c_{1}e^{2t}+c_{2}te^{2t}+2

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y^{''}-4y^'+4y=8

y^{^{''}} - 4 y^{^{'}} + 4 y = 8

y = c_{1} e^{2 t} + c_{2} t e^{2 t} + 2

Solution steps

y^{''} (t) - 4 y^{'} (t) + 4 y = 8

Solve linear ODE:

y = c_{1} e^{2 t} + c_{2} t e^{2 t} + 2

y = c_{1} e^{2 t} + c_{2} t e^{2 t} + 2

y^{^{''}} - y^{^{'}} - 6 y = 14 sin (3 x)

\frac{d ^{2} y}{d x ^{2}} + 2 \frac{d y}{d x} - 15 y = 7 + 4 x

y^{^{''}} + 4 y = - 5 sin (2 t) + 2 cos (2 t)

\frac{d ^{2} y}{d x ^{2}} - 2 \frac{d y}{d x} = 2 e^{2 x}

\frac{d ^{2} y}{d t ^{2}} + 8 \frac{d y}{d t} + 15 y = 3

What is the solution for y^{''}-4y^'+4y=8 ?
The solution for y^{''}-4y^'+4y=8 is y=c_{1}e^{2t}+c_{2}te^{2t}+2