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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

x \to \infty lim (2 + x)

\int_{0}^{1} \frac{1}{x ln ( x )} d x

y = e^{3 e^{x}}

f (x) = arctan (cos (x))

\frac{d}{d x} (4 x^{6} - 3 x^{5} - 10 x^{2} + 5 x + 16)

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