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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''''}} - a^{4} y = 0

u^{^{''}} - a^{2} H (t) = 0

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

y^{^{'''}} + 5 y^{^{''}} - y^{^{'}} - 5 y = 0

y^{^{''}} + 2 y^{^{'}} = 3 y

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