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

The solution for y^{''}-n^2y=(2cos(pin)x)/n is y=c_{1}e^{nx}+c_{2}e^{-nx}-(2(-1)^nx)/(n^3)

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y^{''}-n^2y=(2cos(pin)x)/n

y^{^{''}} - n^{2} y = \frac{2 cos ( πn ) x}{n}

y = c_{1} e^{n x} + c_{2} e^{- n x} - \frac{2 ( - 1 ) ^{n} x}{n ^{3}}

Solution steps

y^{''} (x) - n^{2} y = \frac{2 cos ( πn ) x}{n}

Solve linear ODE:

y = c_{1} e^{n x} + c_{2} e^{- n x} - \frac{2 ( - 1 ) ^{n} x}{n ^{3}}

y = c_{1} e^{n x} + c_{2} e^{- n x} - \frac{2 ( - 1 ) ^{n} x}{n ^{3}}

y^{^{''}} + 8 y^{^{'}} - 2 y = \frac{5}{e ^{3 t}}

y^{^{''}} + 4 y = sin (x), y (0) = 0, y (1) = 0

(d - 4)^{6} y = 5 e^{4} x

y^{^{''}} - 81 y = - 243 t + 162 t^{2}

5 x^{^{''}} + 10 x^{^{'}} + 30 x = 300

What is the solution for y^{''}-n^2y=(2cos(pin)x)/n ?
The solution for y^{''}-n^2y=(2cos(pin)x)/n is y=c_{1}e^{nx}+c_{2}e^{-nx}-(2(-1)^nx)/(n^3)