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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

f (x) = x - 2 sin (x), at x = 0

x y^{^{''}} - y^{^{'}} = x^{2}

\frac{x - 5}{x ^{2} + 1}

\frac{d}{d x} (\frac{24}{x ^{4}})

\frac{d}{d x} (\frac{1}{5 + 3 x})

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