What is the solution for y^{''}-3y^'+2y=10sin(x) ?

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

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y^{''}-3y^'+2y=10sin(x)

y^{^{''}} - 3 y^{^{'}} + 2 y = 10 sin (x)

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

Solution steps

y^{''} (x) - 3 y^{'} (x) + 2 y = 10 sin (x)

Solve linear ODE:

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

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

y^{^{''}} - 2 y^{^{'}} + y = e^{- 2 x}

3 y^{^{''}} + y^{^{'}} - 2 y = 2 cos (t)

x^{^{''}} - x^{^{'}} - 6 x = 8

y^{^{''}} - y^{^{'}} - 2 y = 3 x + 3

\frac{d ^{2} y}{d x ^{2}} - 3 \frac{d y}{d x} + 4 y = x e^{x}

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