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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\int \frac{1}{x ^{2} x ^{2} - 1} d x

d er i v a t i v e f (x) = cos (8 x^{2} - 16) sin (x^{2})

d er i v a t i v e (\frac{3}{5} x)^{5}

t an g e n t \frac{1}{9 x}

\int \frac{37 x + 5}{( 6 x + 1 ) ( x - 1 )} d x

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