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

The solution for y^{''}-4y^'+4y=t,y(0)=0,y^'(0)=1 is y=-1/4 e^{2t}+(5e^{2t}t)/4+t/4+1/4

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y^{''}-4y^'+4y=t,y(0)=0,y^'(0)=1

y^{^{''}} - 4 y^{^{'}} + 4 y = t, y (0) = 0, y^{^{'}} (0) = 1

y = - \frac{1}{4} e^{2 t} + \frac{5 e ^{2 t} t}{4} + \frac{t}{4} + \frac{1}{4}

Solution steps

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

Solve linear ODE:

y = - \frac{1}{4} e^{2 t} + \frac{5 e ^{2 t} t}{4} + \frac{t}{4} + \frac{1}{4}

y = - \frac{1}{4} e^{2 t} + \frac{5 e ^{2 t} t}{4} + \frac{t}{4} + \frac{1}{4}

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

y^{^{''}} - y^{^{'}} = 4 x + 3

y^{^{''}} - 3 y^{^{'}} + 2 y = x, y^{1} (0) = e^{x}

y^{^{''}} - \frac{4}{9} y = x^{3} - 2

x^{^{''}} - 4 x^{^{'}} + 4 x = 84 t^{2} e^{2 t}

What is the solution for y^{''}-4y^'+4y=t,y(0)=0,y^'(0)=1 ?
The solution for y^{''}-4y^'+4y=t,y(0)=0,y^'(0)=1 is y=-1/4 e^{2t}+(5e^{2t}t)/4+t/4+1/4