What is the solution for y^{''}-8y^'+20=0 ?

The solution for y^{''}-8y^'+20=0 is y=c_{1}e^{8t}+c_{2}+(5t)/2

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y^{''}-8y^'+20=0

y^{^{''}} - 8 y^{^{'}} + 20 = 0

y = c_{1} e^{8 t} + c_{2} + \frac{5 t}{2}

Solution steps

y^{''} (t) - 8 y^{'} (t) + 20 = 0

Solve linear ODE:

y = c_{1} e^{8 t} + c_{2} + \frac{5 t}{2}

y = c_{1} e^{8 t} + c_{2} + \frac{5 t}{2}

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What is the solution for y^{''}-8y^'+20=0 ?
The solution for y^{''}-8y^'+20=0 is y=c_{1}e^{8t}+c_{2}+(5t)/2