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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\int 4 x ln (3 x) d x

(\frac{d y}{d x}) - (\frac{y}{3 x}) + (\frac{x}{3 y}) = 0

\int y e^{t y}

\frac{\partial}{\partial x} (\frac{3 x}{( x + y ) ^{3}})

f (x) = - 2 sin (x), at x = \frac{3 π}{4}

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