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)+c_{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) + c_{2} sin (2 t) + \frac{3}{2}

Solution steps

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

Solve linear ODE:

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

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

\frac{\partial}{\partial t} (x z^{2} t)

\frac{d y}{d x} = \frac{2 x}{y} + \frac{2}{y} + \frac{1}{x y}

y = arcsin (8 x + 1)

n = 1 \sum \infty \frac{- 3 ^{n}}{8 ^{n}}

e^{2 x} - 3 x

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)+c_{2}sin(sqrt(2)t)+3/2