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

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

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

y^{^{''}} - 2 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 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 x^{2} (2 x - 1)^{5} d x

\frac{d}{d x} (cos (x^{6}))

u^{4}

f (x) = (1 - x) (x^{2} - 3)^{2}, at x = 2

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