What is the solution for y^{''}-7y^'+12y=6et ?

The solution for y^{''}-7y^'+12y=6et is y=c_{1}e^{4t}+c_{2}e^{3t}+(et)/2+(7e)/(24)

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y^{''}-7y^'+12y=6et

y^{^{''}} - 7 y^{^{'}} + 12 y = 6 e t

y = c_{1} e^{4 t} + c_{2} e^{3 t} + \frac{e t}{2} + \frac{7 e}{24}

Solution steps

y^{''} (t) - 7 y^{'} (t) + 12 y = 6 e t

Solve linear ODE:

y = c_{1} e^{4 t} + c_{2} e^{3 t} + \frac{e t}{2} + \frac{7 e}{24}

y = c_{1} e^{4 t} + c_{2} e^{3 t} + \frac{e t}{2} + \frac{7 e}{24}

\int \frac{x}{( sin ( x ) ) ^{2}} d x

\int x e^{x} + e^{x} d x

\int 2 sin^{2} (y) d y

\frac{d y}{d x} = \frac{y ^{2} - 1}{x ^{2} - 1}, y (2) = - 1

\frac{d}{d x} (ln^{4} (x^{3} + 2 x - e^{- 3 x}))

What is the solution for y^{''}-7y^'+12y=6et ?
The solution for y^{''}-7y^'+12y=6et is y=c_{1}e^{4t}+c_{2}e^{3t}+(et)/2+(7e)/(24)