What is the solution for y^{''}-10y^'+25y=75t+295 ?

The solution for y^{''}-10y^'+25y=75t+295 is y=c_{1}e^{5t}+c_{2}te^{5t}+3t+13

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y^{''}-10y^'+25y=75t+295

y^{^{''}} - 10 y^{^{'}} + 25 y = 75 t + 295

y = c_{1} e^{5 t} + c_{2} t e^{5 t} + 3 t + 13

Solution steps

y^{''} (t) - 10 y^{'} (t) + 25 y = 75 t + 295

Solve linear ODE:

y = c_{1} e^{5 t} + c_{2} t e^{5 t} + 3 t + 13

y = c_{1} e^{5 t} + c_{2} t e^{5 t} + 3 t + 13

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

y^{^{''}} - 6 y^{^{'}} + 9 y = \frac{1}{x}

y^{^{''}} + 2 y^{^{'}} + 5 y = 3 e^{2 x}, y (0) = 1, y^{^{'}} (0) = 1

y^{^{'''}} - 2 y^{^{''}} + y = 5 + 8 x e^{x}

y^{^{''}} - y^{^{'}} - 2 y = 8 x + 9

What is the solution for y^{''}-10y^'+25y=75t+295 ?
The solution for y^{''}-10y^'+25y=75t+295 is y=c_{1}e^{5t}+c_{2}te^{5t}+3t+13