What is the solution for y^{''}+2y^'+5y=0,y(0)=3,y^'(0)=-2 ?

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

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y^{''}+2y^'+5y=0,y(0)=3,y^'(0)=-2

y^{^{''}} + 2 y^{^{'}} + 5 y = 0, y (0) = 3, y^{^{'}} (0) = - 2

y = e^{- t} (3 cos (2 t) + \frac{1}{2} sin (2 t))

Solution steps

y^{''} (t) + 2 y^{'} (t) + 5 y = 0

Solve linear ODE:

y = e^{- t} (3 cos (2 t) + \frac{1}{2} sin (2 t))

y = e^{- t} (3 cos (2 t) + \frac{1}{2} sin (2 t))

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

8 y^{^{''}} - 8 y^{^{'}} + 20 y = 0

y^{^{''}} + 4 y^{^{'}} + 2 y = 0, y (0) = 0, y^{^{'}} (0) = 1

\frac{d ^{2} y}{d t ^{2}} + 16 y = 0

\frac{d ^{2} x}{d t ^{2}} + 2 \frac{d x}{d t} + 8 x = 0

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