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

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

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

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

y = \frac{3}{4} e^{2 t} - \frac{3 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{3}{4} e^{2 t} - \frac{3 e ^{2 t} t}{4} + \frac{t}{4} + \frac{1}{4}

y = \frac{3}{4} e^{2 t} - \frac{3 e ^{2 t} t}{4} + \frac{t}{4} + \frac{1}{4}

(D^{2} + 4 D + 5) y = 50 x + 13 e^{x}

(d^{2} - 4) y (d, t) = 5 e^{3 t}, y (0) = 1

y^{^{''}} - 3 y^{^{'}} = 2 e^{x} sin (x)

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

y^{^{''}} + 2 y^{^{'}} + 3 y = 12 e^{x} + 3 x

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