What is the solution for y^{''}-y^'=e^tcos(t) ?

The solution for y^{''}-y^'=e^tcos(t) is y=c_{1}e^t+c_{2}+1/2 e^tsin(t)-1/2 e^tcos(t)

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y^{''}-y^'=e^tcos(t)

y^{^{''}} - y^{^{'}} = e^{t} cos (t)

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

Solution steps

y^{''} (t) - y^{'} (t) = e^{t} cos (t)

Solve linear ODE:

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

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

\int \frac{1}{e ^{x} - e ^{- x}} d x

\int sin (x n) d x

\frac{d P}{d t} = \frac{9 P}{900} (9 - P), P (0) = 4

- 12 sin (2 x)

\int x^{3} \cdot 9 - x^{2} d x

What is the solution for y^{''}-y^'=e^tcos(t) ?
The solution for y^{''}-y^'=e^tcos(t) is y=c_{1}e^t+c_{2}+1/2 e^tsin(t)-1/2 e^tcos(t)