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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\int cos (a) x d x

d er i v a t i v e x e^{y}

\int (9 - x^{2}) d x

a re a \frac{( x ^{2} )}{2}, x + 4

d er i v a t i v e f (θ) = sin (θ) cos (θ)

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