What is the solution for y^{''}-y=5sin(2t)(0)1 ?

The solution for y^{''}-y=5sin(2t)(0)1 is y=c_{1}e^t+c_{2}e^{-t}

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y^{''}-y=5sin(2t)(0)1

y^{^{''}} - y = 5 sin (2 t) (0) 1

y = c_{1} e^{t} + c_{2} e^{- t}

Solution steps

y^{''} (t) - y = 5 sin (2 t) (0) 1

Solve linear ODE:

y = c_{1} e^{t} + c_{2} e^{- t}

y = c_{1} e^{t} + c_{2} e^{- t}

\frac{1}{2 + 3 x}

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

\frac{r}{r ^{2} + 1}

\int (- 0.5 x + 450) d x

\frac{d y}{d x} = 4 x - 5

What is the solution for y^{''}-y=5sin(2t)(0)1 ?
The solution for y^{''}-y=5sin(2t)(0)1 is y=c_{1}e^t+c_{2}e^{-t}