What is the solution for y^{''}-8y^'+16y=t,y(0)=0,y^'(0)=1 ?

The solution for y^{''}-8y^'+16y=t,y(0)=0,y^'(0)=1 is y=-1/32 e^{4t}+(17e^{4t}t)/(16)+t/(16)+1/32

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y^{''}-8y^'+16y=t,y(0)=0,y^'(0)=1

y^{^{''}} - 8 y^{^{'}} + 16 y = t, y (0) = 0, y^{^{'}} (0) = 1

y = - \frac{1}{32} e^{4 t} + \frac{17 e ^{4 t} t}{16} + \frac{t}{16} + \frac{1}{32}

Solution steps

y^{''} (t) - 8 y^{'} (t) + 16 y = t

Solve linear ODE:

y = - \frac{1}{32} e^{4 t} + \frac{17 e ^{4 t} t}{16} + \frac{t}{16} + \frac{1}{32}

y = - \frac{1}{32} e^{4 t} + \frac{17 e ^{4 t} t}{16} + \frac{t}{16} + \frac{1}{32}

What is the solution for y^{''}-8y^'+16y=t,y(0)=0,y^'(0)=1 ?
The solution for y^{''}-8y^'+16y=t,y(0)=0,y^'(0)=1 is y=-1/32 e^{4t}+(17e^{4t}t)/(16)+t/(16)+1/32