What is the solution for y^{''}-6y^'+8y=e^{3x}sin(4x) ?

The solution for y^{''}-6y^'+8y=e^{3x}sin(4x) is y=c_{1}e^{4x}+c_{2}e^{2x}-1/17 e^{3x}sin(4x)

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y^{''}-6y^'+8y=e^{3x}sin(4x)

y^{^{''}} - 6 y^{^{'}} + 8 y = e^{3 x} sin (4 x)

y = c_{1} e^{4 x} + c_{2} e^{2 x} - \frac{1}{17} e^{3 x} sin (4 x)

Solution steps

y^{''} (x) - 6 y^{'} (x) + 8 y = e^{3 x} sin (4 x)

Solve linear ODE:

y = c_{1} e^{4 x} + c_{2} e^{2 x} - \frac{1}{17} e^{3 x} sin (4 x)

y = c_{1} e^{4 x} + c_{2} e^{2 x} - \frac{1}{17} e^{3 x} sin (4 x)

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

y^{^{''}} + 4 y = 2 t

y^{^{''}} + 9 y = cos (2 x)

y^{^{''}} - 10 y^{^{'}} + 25 y = x

d (x) = - 3 x^{2} - 15 x + 146

What is the solution for y^{''}-6y^'+8y=e^{3x}sin(4x) ?
The solution for y^{''}-6y^'+8y=e^{3x}sin(4x) is y=c_{1}e^{4x}+c_{2}e^{2x}-1/17 e^{3x}sin(4x)