What is the solution for y^{''}-3y^'+2y=t ?

The solution for y^{''}-3y^'+2y=t is y=c_{1}e^{2t}+c_{2}e^t+t/2+3/4

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y^{''}-3y^'+2y=t

y^{^{''}} - 3 y^{^{'}} + 2 y = t

y = c_{1} e^{2 t} + c_{2} e^{t} + \frac{t}{2} + \frac{3}{4}

Solution steps

y^{''} (t) - 3 y^{'} (t) + 2 y = t

Solve linear ODE:

y = c_{1} e^{2 t} + c_{2} e^{t} + \frac{t}{2} + \frac{3}{4}

y = c_{1} e^{2 t} + c_{2} e^{t} + \frac{t}{2} + \frac{3}{4}

\frac{d}{d y} (sin (x + y))

x \to π lim (x^{3} cos (x))

G (t) = \frac{1 - 3 t}{5 + t}

\frac{d}{d x} ((A x^{2} + B x + C) e^{3 x})

\frac{d y}{d x} = x e^{- 2 x}

What is the solution for y^{''}-3y^'+2y=t ?
The solution for y^{''}-3y^'+2y=t is y=c_{1}e^{2t}+c_{2}e^t+t/2+3/4