What is the solution for y^{''}-2*y^'=sin(x) ?

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

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y^{''}-2*y^'=sin(x)

y^{^{''}} - 2 \cdot y^{^{'}} = sin (x)

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

Solution steps

y^{''} (x) - 2 \cdot y^{'} (x) = sin (x)

Solve linear ODE:

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

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

\int_{0}^{24} 9 + 3 x d x

\frac{d ^{2}}{d x ^{2}} ((x^{2} + 6)^{9})

\frac{d}{d x} ((x - 1)^{n})

y = 9 - 2 x, y = x^{2} - 39

e^{y} y^{^{'}} = (x + e^{y} - 1)

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