What is the solution for y^{''}-5y^'=2t ?

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

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y^{''}-5y^'=2t

y^{^{''}} - 5 y^{^{'}} = 2 t

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

Solution steps

y^{''} (t) - 5 y^{'} (t) = 2 t

Solve linear ODE:

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

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

f (x) = 2 x^{2} + x - 1

n = 4 \sum \infty n! (x - 4)^{n}

3 x e^{2 x}

t^{2} \frac{d y}{d t} + t y = 3

\frac{d}{d t} (5 t + \frac{7}{t})

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