What is the solution for y^{''}+4y=(5x^2+4x-3)e^x ?

The solution for y^{''}+4y=(5x^2+4x-3)e^x is y=c_{1}cos(2x)+c_{2}sin(2x)+e^xx^2-e^x

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y^{''}+4y=(5x^2+4x-3)e^x

y^{^{''}} + 4 y = (5 x^{2} + 4 x - 3) e^{x}

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

Solution steps

y^{''} (x) + 4 y = (5 x^{2} + 4 x - 3) e^{x}

Solve linear ODE:

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

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

(\frac{1}{2}) y^{^{''}} + y^{^{'}} + 5 y = sin (3 t)

y^{^{''''}} + 25 y^{^{''}} = sin (5 t) + t e^{6 t} + 4

6 y^{^{''}} - 10 y^{^{'}} - 4 y = x + 2

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

y^{^{''}} - y^{^{'}} = x e [x], y (0) = 2, y^{^{'}} (0) = 1

What is the solution for y^{''}+4y=(5x^2+4x-3)e^x ?
The solution for y^{''}+4y=(5x^2+4x-3)e^x is y=c_{1}cos(2x)+c_{2}sin(2x)+e^xx^2-e^x