What is the solution for y^{''}-8y^'+16y=t-2e^{4t} ?

The solution for y^{''}-8y^'+16y=t-2e^{4t} is y=c_{1}e^{4t}+c_{2}te^{4t}+t/(16)+1/32-e^{4t}t^2

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y^{''}-8y^'+16y=t-2e^{4t}

y^{^{''}} - 8 y^{^{'}} + 16 y = t - 2 e^{4 t}

y = c_{1} e^{4 t} + c_{2} t e^{4 t} + \frac{t}{16} + \frac{1}{32} - e^{4 t} t^{2}

Solution steps

y^{''} (t) - 8 y^{'} (t) + 16 y = t - 2 e^{4 t}

Solve linear ODE:

y = c_{1} e^{4 t} + c_{2} t e^{4 t} + \frac{t}{16} + \frac{1}{32} - e^{4 t} t^{2}

y = c_{1} e^{4 t} + c_{2} t e^{4 t} + \frac{t}{16} + \frac{1}{32} - e^{4 t} t^{2}

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\int_{0}^{1} \frac{31}{4 y - 1} d y

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What is the solution for y^{''}-8y^'+16y=t-2e^{4t} ?
The solution for y^{''}-8y^'+16y=t-2e^{4t} is y=c_{1}e^{4t}+c_{2}te^{4t}+t/(16)+1/32-e^{4t}t^2