What is the solution for y^{''}+4y=xe^x ?

The solution for y^{''}+4y=xe^x is y=c_{1}cos(2x)+c_{2}sin(2x)+(e^xx)/5-2/25 e^x

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y^{''}+4y=xe^x

y^{^{''}} + 4 y = x e^{x}

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{e ^{x} x}{5} - \frac{2}{25} e^{x}

Solution steps

y^{''} (x) + 4 y = x e^{x}

Solve linear ODE:

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{e ^{x} x}{5} - \frac{2}{25} e^{x}

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{e ^{x} x}{5} - \frac{2}{25} e^{x}

y^{^{''}} - 0.2 y^{^{'}} + 0.01 y = 0, y (0) = 14, y^{^{'}} (0) = b

\int sin (3 θ) cos (5 θ) d θ

l a pl a ce t r an s f or m 2 (t - 2)^{2} + 7 (t - 2) + 9

\frac{d}{d x} (sin (2 x^{3} + x^{2} + 1))

x \to 0 lim (\frac{a x - a ln ( x + 1 )}{x ln ( x + 1 )})

What is the solution for y^{''}+4y=xe^x ?
The solution for y^{''}+4y=xe^x is y=c_{1}cos(2x)+c_{2}sin(2x)+(e^xx)/5-2/25 e^x