What is the solution for y^{''}-8y^'*+25=0 ?

The solution for y^{''}-8y^'*+25=0 is y=c_{1}e^{8t}+c_{2}+(25t)/8

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y^{''}-8y^'*+25=0

y^{^{''}} - 8 y^{^{'}} \cdot + 25 = 0

y = c_{1} e^{8 t} + c_{2} + \frac{25 t}{8}

Solution steps

y^{''} (t) - 8 y^{'} (t) \cdot + 25 = 0

Solve linear ODE:

y = c_{1} e^{8 t} + c_{2} + \frac{25 t}{8}

y = c_{1} e^{8 t} + c_{2} + \frac{25 t}{8}

y^{^{''}} + 4 y^{^{'}} - 5 y = 75 t^{3}

y^{^{''}} - 7 y^{^{'}} + 12 y = 5 sin (x)

y^{^{''}} + 3 y^{^{'}} + 2 y = 18 + e^{x}

y^{^{''}} + 49 y = cos (7 t)

y^{^{''}} - 9 y = 12 cos (3 x)

What is the solution for y^{''}-8y^'*+25=0 ?
The solution for y^{''}-8y^'*+25=0 is y=c_{1}e^{8t}+c_{2}+(25t)/8