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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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f (x) = \frac{3}{x}, (- 5, - \frac{3}{5})

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