What is the solution for y^{''}-5y^'=8t ?

The solution for y^{''}-5y^'=8t is y=c_{1}e^{5t}+c_{2}-(4t^2)/5-(8t)/(25)

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y^{''}-5y^'=8t

y^{^{''}} - 5 y^{^{'}} = 8 t

y = c_{1} e^{5 t} + c_{2} - \frac{4 t ^{2}}{5} - \frac{8 t}{25}

Solution steps

y^{''} (t) - 5 y^{'} (t) = 8 t

Solve linear ODE:

y = c_{1} e^{5 t} + c_{2} - \frac{4 t ^{2}}{5} - \frac{8 t}{25}

y = c_{1} e^{5 t} + c_{2} - \frac{4 t ^{2}}{5} - \frac{8 t}{25}

y^{^{'}} = y, y (0) = 1

\int \frac{( 2 x + 1 )}{x ^{2} - 1} d x

2 e^{x} (cos (x) - sin (x))

\int_{- 2}^{3} (x^{3} - x^{2} - 6 x) d x

\frac{d}{d t} (arcsin (t))

What is the solution for y^{''}-5y^'=8t ?
The solution for y^{''}-5y^'=8t is y=c_{1}e^{5t}+c_{2}-(4t^2)/5-(8t)/(25)