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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 16 y = 10 cos (4 x) + 15 sin (4 x)

y^{^{''}} - 10 y^{^{'}} + 24 y = 48 x^{2} + 10

y^{^{''}} + 4 y = t^{2} + 10 e^{t}, y (0) = 0, y^{^{'}} (0) = 3

y^{^{''}} + y = 4 x \cdot sin (x)

y^{^{''}} - 10 y^{^{'}} + 25 y = \frac{1.5 e ^{5 t}}{t ^{2} + 1}

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