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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

n = 0 \sum \infty x^{3 n}

\int 24 e^{20 t} d t

x \to - 3 - lim (\frac{2}{x ^{2} - 9})

y = x^{5} + cos^{5} (x)

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