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

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

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

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

y = c_{1} + c_{2} e^{- x} + sin (x) - cos (x)

Solution steps

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

Solve linear ODE:

y = c_{1} + c_{2} e^{- x} + sin (x) - cos (x)

y = c_{1} + c_{2} e^{- x} + sin (x) - cos (x)

y^{^{''}} + y^{^{'}} - 2 y = 2 x + 3, y (0) = 1, y^{^{'}} (0) = - 4

y^{^{''}} + 5 y^{^{'}} + 4 y = 10 t e^{5 t}

\frac{d ^{2} y}{d x ^{2}} - 3 \frac{d y}{d x} + 2 = 0

y^{^{''}} + 2 y^{^{'}} + y = e^{- t} \cdot ln (t)

y^{^{''}} + 9 y = 18 x + 36 sin (x)

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