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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\int \frac{4}{1 + x ^{2}} d x

3 y^{^{'}} - 2 y = e^{- \frac{π}{2}}

y = x^{2 c o s (x)}

\frac{d}{d x} (3 - ∣ x - 3 ∣)

n = 0 \sum \infty 7 (- \frac{1}{8})^{n}

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