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Popular Calculus >

y^{''}-2y^'-8y=0,y(1)=1,y^'(1)=0

Solution

y^{^{''}} - 2 y^{^{'}} - 8 y = 0, y (1) = 1, y^{^{'}} (1) = 0

Solution

y = e^{4 t - 6} (e^{2} - \frac{2 e ^{2}}{3}) + \frac{2 e ^{2}}{3} e^{- 2 t}

Solution steps

y^{''} (t) - 2 y^{'} (t) - 8 y = 0

Solve linear ODE:

y = e^{4 t - 6} (e^{2} - \frac{2 e ^{2}}{3}) + \frac{2 e ^{2}}{3} e^{- 2 t}

y = e^{4 t - 6} (e^{2} - \frac{2 e ^{2}}{3}) + \frac{2 e ^{2}}{3} e^{- 2 t}

Graph

Popular Examples

y^{''}+16y=0,y(pi)=0,y(0)=0

y^{^{''}} + 16 y = 0, y (π) = 0, y (0) = 0

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

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

y^{''}+8y^'+16y=0,y(0)=6,y^'(0)=2

y^{^{''}} + 8 y^{^{'}} + 16 y = 0, y (0) = 6, y^{^{'}} (0) = 2

y^{''}+121y=0,y(0)=sin(kt)

y^{^{''}} + 121 y = 0, y (0) = sin (k t)

2y^{''}+11y^'+8y=0,y(0)=5,y^'(0)=0

2 y^{^{''}} + 11 y^{^{'}} + 8 y = 0, y (0) = 5, y^{^{'}} (0) = 0

Frequently Asked Questions (FAQ)

What is the solution for y^{''}-2y^'-8y=0,y(1)=1,y^'(1)=0 ?
The solution for y^{''}-2y^'-8y=0,y(1)=1,y^'(1)=0 is y=e^{4t-6}(e^2-(2e^2)/3)+(2e^2)/3 e^{-2t}