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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - y = sin (t), y (0) = 0, y^{^{'}} (0) = 1

y^{^{''}} + 6 y^{^{'}} + 9 y = 2 sin (x)

y^{^{''}} - 8 y^{^{'}} + 16 y = 8 e^{4 x}

y^{^{''}} + 2 y^{^{'}} + y = \frac{e ^{- x}}{x ^{2} + 1}

y^{^{''}} - 2 y^{^{'}} + 17 y = e^{x} cos (4 x)

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