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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

(d^{2} - d - 2) y = 6 e^{\frac{- t}{2}}

d, d (q) = - 0.4 q + 23

(d^{2} + 4) y = 4 sin^{2} (x)

(d^{4} + 2 d^{3} - 3 d^{2}) y = 3 e^{2} x + 4 sin (x)

(d^{2} - 4) y = 3 e^{2} x - 4 e^{- 2} x

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