What is the solution for y^{''}-3y^'+2y=sin(x)+e^x ?

The solution for y^{''}-3y^'+2y=sin(x)+e^x is y=c_{1}e^{2x}+c_{2}e^x+1/10 sin(x)+3/10 cos(x)-e^xx

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y^{''}-3y^'+2y=sin(x)+e^x

y^{^{''}} - 3 y^{^{'}} + 2 y = sin (x) + e^{x}

y = c_{1} e^{2 x} + c_{2} e^{x} + \frac{1}{10} sin (x) + \frac{3}{10} cos (x) - e^{x} x

Solution steps

y^{''} (x) - 3 y^{'} (x) + 2 y = sin (x) + e^{x}

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} e^{x} + \frac{1}{10} sin (x) + \frac{3}{10} cos (x) - e^{x} x

y = c_{1} e^{2 x} + c_{2} e^{x} + \frac{1}{10} sin (x) + \frac{3}{10} cos (x) - e^{x} x

5 y^{^{''}} + 25 y^{^{'}} + \frac{1}{1 0 ^{- 2}} y = 10 e^{- \frac{x}{4}}

y^{^{''}} + 2 y^{^{'}} - 35 y = 13 sin (x) - e^{3 x} + 1

y^{^{''}} + a y^{^{'}} + y = e^{- 4 t}

y^{^{''}} + 9 y = 6 cos (3 x)

(D^{2} - 3 D + 2) y = e^{3 x}

What is the solution for y^{''}-3y^'+2y=sin(x)+e^x ?
The solution for y^{''}-3y^'+2y=sin(x)+e^x is y=c_{1}e^{2x}+c_{2}e^x+1/10 sin(x)+3/10 cos(x)-e^xx