What is the solution for 4y^{''}-40y^'+96y=5sin(2x) ?

The solution for 4y^{''}-40y^'+96y=5sin(2x) is y=c_{1}e^{6x}+c_{2}e^{4x}+1/32 sin(2x)+1/32 cos(2x)

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4y^{''}-40y^'+96y=5sin(2x)

4 y^{^{''}} - 40 y^{^{'}} + 96 y = 5 sin (2 x)

y = c_{1} e^{6 x} + c_{2} e^{4 x} + \frac{1}{32} sin (2 x) + \frac{1}{32} cos (2 x)

Solution steps

4 y^{''} (x) - 40 y^{'} (x) + 96 y = 5 sin (2 x)

Solve linear ODE:

y = c_{1} e^{6 x} + c_{2} e^{4 x} + \frac{1}{32} sin (2 x) + \frac{1}{32} cos (2 x)

y = c_{1} e^{6 x} + c_{2} e^{4 x} + \frac{1}{32} sin (2 x) + \frac{1}{32} cos (2 x)

\int_{1}^{e^{2}} x^{3} ln (x) d x

x + (\frac{1}{x})

f (x) = x^{2} - 4

\frac{d}{d x} (\frac{1}{2 - cos ( π ) x})

x \to - 6 + lim (\frac{1}{( x + 6 ) ^{4}})

What is the solution for 4y^{''}-40y^'+96y=5sin(2x) ?
The solution for 4y^{''}-40y^'+96y=5sin(2x) is y=c_{1}e^{6x}+c_{2}e^{4x}+1/32 sin(2x)+1/32 cos(2x)