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

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

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

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

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{1}{3} sin (x)

Solution steps

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

Solve linear ODE:

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{1}{3} sin (x)

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{1}{3} sin (x)

\frac{2 x ^{2} + 1}{x ^{2} + 8}

\int_{0}^{1} ln (x^{2} + 1) d x

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

- \frac{3 ( z ^{2} + 1 )}{z ^{2}}

y^{2} = 6 - x, 3 y = x + 12

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