What is the solution for y^{''}-2y^'=sin(x),y(0)=1,y^'(0)=2 ?

The solution for y^{''}-2y^'=sin(x),y(0)=1,y^'(0)=2 is y=(1/2+3/5)e^{2x}-1/2-1/5 sin(x)+2/5 cos(x)

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y^{''}-2y^'=sin(x),y(0)=1,y^'(0)=2

y^{^{''}} - 2 y^{^{'}} = sin (x), y (0) = 1, y^{^{'}} (0) = 2

y = (\frac{1}{2} + \frac{3}{5}) e^{2 x} - \frac{1}{2} - \frac{1}{5} sin (x) + \frac{2}{5} cos (x)

Solution steps

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

Solve linear ODE:

y = (\frac{1}{2} + \frac{3}{5}) e^{2 x} - \frac{1}{2} - \frac{1}{5} sin (x) + \frac{2}{5} cos (x)

y = (\frac{1}{2} + \frac{3}{5}) e^{2 x} - \frac{1}{2} - \frac{1}{5} sin (x) + \frac{2}{5} cos (x)

\int t^{2} + 4 d t

\frac{d}{d x} (sin (x) + \frac{5}{6} cot (x))

\int_{- 8}^{0} (4 - 9 x)^{- 0.5} d x

(x - 1)^{3}, (x - 1)

\frac{\partial}{\partial x} (\frac{1}{2} ln (x^{2} d))

What is the solution for y^{''}-2y^'=sin(x),y(0)=1,y^'(0)=2 ?
The solution for y^{''}-2y^'=sin(x),y(0)=1,y^'(0)=2 is y=(1/2+3/5)e^{2x}-1/2-1/5 sin(x)+2/5 cos(x)