What is the solution for y^{''}+4y=2e^x+sin(2x) ?

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

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y^{''}+4y=2e^x+sin(2x)

y^{^{''}} + 4 y = 2 e^{x} + sin (2 x)

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{2}{5} e^{x} - \frac{x cos ( 2 x )}{4}

Solution steps

y^{''} (x) + 4 y = 2 e^{x} + sin (2 x)

Solve linear ODE:

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{2}{5} e^{x} - \frac{x cos ( 2 x )}{4}

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{2}{5} e^{x} - \frac{x cos ( 2 x )}{4}

\frac{\partial}{\partial y} (\frac{x ^{2} + y ^{2}}{x ^{2} - y ^{2}})

\frac{\partial}{\partial x} (4 - x^{2} - 4 Y (x)^{2})

\frac{d}{d x} (- x^{3} + 5 x^{2} - 6 x)

\int \frac{( x ^{2} )}{1 + x ^{2}} d x

y = \frac{4}{5 - 3 ( x )}

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