What is the solution for y^{''}+12y=8e^{4x} ?

The solution for y^{''}+12y=8e^{4x} is y=c_{1}cos(2sqrt(3)x)+c_{2}sin(2sqrt(3)x)+2/7 e^{4x}

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y^{''}+12y=8e^{4x}

y^{^{''}} + 12 y = 8 e^{4 x}

y = c_{1} cos (23 x) + c_{2} sin (23 x) + \frac{2}{7} e^{4 x}

Solution steps

y^{''} (x) + 12 y = 8 e^{4 x}

Solve linear ODE:

y = c_{1} cos (23 x) + c_{2} sin (23 x) + \frac{2}{7} e^{4 x}

y = c_{1} cos (23 x) + c_{2} sin (23 x) + \frac{2}{7} e^{4 x}

y^{^{''}} + 2 y^{^{'}} + y = \frac{t}{e ^{t}}

y^{^{''}} + y = 2 sin (2 x)

y^{^{'''}} + 49 y^{^{'}} = 5 x + 6

y^{^{''}} - 6 y^{^{'}} + 9 y = x^{- 4} \cdot e^{3} x

3 y^{^{''}} + 4 y = 3 e^{3} x

What is the solution for y^{''}+12y=8e^{4x} ?
The solution for y^{''}+12y=8e^{4x} is y=c_{1}cos(2sqrt(3)x)+c_{2}sin(2sqrt(3)x)+2/7 e^{4x}