What is the solution for y^{''}+2y=3,y(0)=-1,y^'(0)=2 ?

The solution for y^{''}+2y=3,y(0)=-1,y^'(0)=2 is y=-5/2 cos(sqrt(2)t)+sqrt(2)sin(sqrt(2)t)+3/2

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

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

y = - \frac{5}{2} cos (2 t) + 2 sin (2 t) + \frac{3}{2}

Solution steps

y^{''} (t) + 2 y = 3

Solve linear ODE:

y = - \frac{5}{2} cos (2 t) + 2 sin (2 t) + \frac{3}{2}

y = - \frac{5}{2} cos (2 t) + 2 sin (2 t) + \frac{3}{2}

\frac{d}{d x} (lo g_{5} (2 x^{3} + x^{2}))

e^{x} y \frac{d y}{d x} = e^{- y} + e^{- 8 x - y}

\frac{\partial}{\partial x} ((5 y^{4}) cos (5 x))

d er i v a t i v e h (t) = (t + 2)^{\frac{2}{3}} (2 t^{2} - 3)^{3}

\frac{d y}{d t} = (- csc (t) cot (t)) e^{- c s c (t)}

What is the solution for y^{''}+2y=3,y(0)=-1,y^'(0)=2 ?
The solution for y^{''}+2y=3,y(0)=-1,y^'(0)=2 is y=-5/2 cos(sqrt(2)t)+sqrt(2)sin(sqrt(2)t)+3/2