What is the solution for y^{''}-y=18e^{5x} ?

The solution for y^{''}-y=18e^{5x} is y=c_{1}e^x+c_{2}e^{-x}+3/4 e^{5x}

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y^{''}-y=18e^{5x}

y^{^{''}} - y = 18 e^{5 x}

y = c_{1} e^{x} + c_{2} e^{- x} + \frac{3}{4} e^{5 x}

Solution steps

y^{''} (x) - y = 18 e^{5 x}

Solve linear ODE:

y = c_{1} e^{x} + c_{2} e^{- x} + \frac{3}{4} e^{5 x}

y = c_{1} e^{x} + c_{2} e^{- x} + \frac{3}{4} e^{5 x}

y^{^{''}} + 4 y^{^{'}} + 8 y = 160

- x^{^{''}} + 4 x + 4 = 0

y^{^{''}} - 6 y^{^{'}} + 45 y = 2 sin (3 t)

\frac{d ^{2} y}{d x ^{2}} (1) + \frac{d y}{d x} (1) = tan (x)

y^{^{''}} + 4 y = sin (x) \cdot sin (2 x)

What is the solution for y^{''}-y=18e^{5x} ?
The solution for y^{''}-y=18e^{5x} is y=c_{1}e^x+c_{2}e^{-x}+3/4 e^{5x}