What is the solution for y^{''}-4y^'+4y=1,y^'(0)=0 ?

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

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y^{''}-4y^'+4y=1,y^'(0)=0

y^{^{''}} - 4 y^{^{'}} + 4 y = 1, y^{^{'}} (0) = 0

y = - \frac{c _{2} e ^{2 t}}{2} + c_{2} e^{2 t} t + \frac{1}{4}

Solution steps

y^{''} (t) - 4 y^{'} (t) + 4 y = 1

Solve linear ODE:

y = - \frac{c _{2} e ^{2 t}}{2} + c_{2} e^{2 t} t + \frac{1}{4}

y = - \frac{c _{2} e ^{2 t}}{2} + c_{2} e^{2 t} t + \frac{1}{4}

y^{^{''}} - 10 y^{^{'}} + 21 y = 240 e^{t}

20 \frac{d ^{4} y}{d x ^{4}} + 100 \frac{d ^{2} y}{d x ^{2}} = 139

y^{^{''}} - y = 1, y^{^{'}} (0) = 1

y^{^{''}} + 2 y^{^{'}} - 3 y = 6 x + 5

5 y^{^{''}} + 4 y^{^{'}} + 4 y = 35 e^{- 4 x}

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