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Popular >

y^{''}+32y^'+192y=0,y(0)= 1/6 ,y^'(0)=2

Solution

y^{^{''}} + 32 y^{^{'}} + 192 y = 0, y (0) = \frac{1}{6}, y^{^{'}} (0) = 2

Solution

y = (\frac{1}{6} + \frac{5}{24}) e^{- 8 t} - \frac{5}{24} e^{- 24 t}

Solution steps

y^{''} (t) + 32 y^{'} (t) + 192 y = 0

Solve linear ODE:

y = (\frac{1}{6} + \frac{5}{24}) e^{- 8 t} - \frac{5}{24} e^{- 24 t}

y = (\frac{1}{6} + \frac{5}{24}) e^{- 8 t} - \frac{5}{24} e^{- 24 t}

Graph

Popular Examples

y^{''}+15y^'+50y=0,y(0)=12,y^'(0)=-100

y^{^{''}} + 15 y^{^{'}} + 50 y = 0, y (0) = 12, y^{^{'}} (0) = - 100

y^{'''''}+4y^{'''}+4y^'=0

y^{^{'''''}} + 4 y^{^{'''}} + 4 y^{^{'}} = 0

y^{''}+15y^'+50y=0,y(0)=12,y^'(0)=-105

y^{^{''}} + 15 y^{^{'}} + 50 y = 0, y (0) = 12, y^{^{'}} (0) = - 105

x^{''}+x^'+0.25x=0,x(0)=2,x^'(0)= 1/3

x^{^{''}} + x^{^{'}} + 0.25 x = 0, x (0) = 2, x^{^{'}} (0) = \frac{1}{3}

y^{''}+9y=0,y(pi/6)=1,y(0)=-1

y^{^{''}} + 9 y = 0, y (\frac{π}{6}) = 1, y (0) = - 1

Frequently Asked Questions (FAQ)

What is the solution for y^{''}+32y^'+192y=0,y(0)= 1/6 ,y^'(0)=2 ?
The solution for y^{''}+32y^'+192y=0,y(0)= 1/6 ,y^'(0)=2 is y=(1/6+5/24)e^{-8t}-5/24 e^{-24t}