What is the solution for y^{''}-9y=cos^2(x) ?

The solution for y^{''}-9y=cos^2(x) is y=c_{1}e^{3x}+c_{2}e^{-3x}+(-26cos^2(x)-5cos(2x)-13)/(468)

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y^{''}-9y=cos^2(x)

y^{^{''}} - 9 y = cos^{2} (x)

y = c_{1} e^{3 x} + c_{2} e^{- 3 x} + \frac{- 26 cos ^{2} ( x ) - 5 cos ( 2 x ) - 13}{468}

Solution steps

y^{''} (x) - 9 y = cos^{2} (x)

Solve linear ODE:

y = c_{1} e^{3 x} + c_{2} e^{- 3 x} + \frac{- 26 cos ^{2} ( x ) - 5 cos ( 2 x ) - 13}{468}

y = c_{1} e^{3 x} + c_{2} e^{- 3 x} + \frac{- 26 cos ^{2} ( x ) - 5 cos ( 2 x ) - 13}{468}

y^{^{''}} - y^{^{'}} - 6 y = e^{3 t} + 5 t

y^{^{''}} + y^{^{'}} - 3 y = 1

y^{^{''}} + 3 y^{^{'}} + 2 y = 2 cos^{2} (x)

y^{^{''}} - 24 y = 2 x cos (5 x)

\frac{d ^{2} y}{d x ^{2}} - 3 \frac{d y}{d x} + 2 y = 2 x^{2} + e^{x} + 2 x e^{x} + 4 e^{3 x}

What is the solution for y^{''}-9y=cos^2(x) ?
The solution for y^{''}-9y=cos^2(x) is y=c_{1}e^{3x}+c_{2}e^{-3x}+(-26cos^2(x)-5cos(2x)-13)/(468)