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Popular >

4y^{''}+12y^'+9y=0,y(0)=1,y^'(0)=-4

Solution

4 y^{^{''}} + 12 y^{^{'}} + 9 y = 0, y (0) = 1, y^{^{'}} (0) = - 4

Solution

y = e^{- \frac{3 t}{2}} - \frac{5 e ^{- \frac{3 t}{2}} t}{2}

Solution steps

4 y^{''} (t) + 12 y^{'} (t) + 9 y = 0

Solve linear ODE:

y = e^{- \frac{3 t}{2}} - \frac{5 e ^{- \frac{3 t}{2}} t}{2}

y = e^{- \frac{3 t}{2}} - \frac{5 e ^{- \frac{3 t}{2}} t}{2}

Graph

Popular Examples

y^{''}+5y^'=0,y(0)=5,y^'(0)=-20

y^{^{''}} + 5 y^{^{'}} = 0, y (0) = 5, y^{^{'}} (0) = - 20

y^{''}+4y^'+7y=0,y(0)=1,y^'(0)=-2

y^{^{''}} + 4 y^{^{'}} + 7 y = 0, y (0) = 1, y^{^{'}} (0) = - 2

y^{''}-2y^'-3y=0,y(0)=0,y^'(0)=-4

y^{^{''}} - 2 y^{^{'}} - 3 y = 0, y (0) = 0, y^{^{'}} (0) = - 4

y^{''}+81y=0,y(0)=0,y^'(0)=3

y^{^{''}} + 81 y = 0, y (0) = 0, y^{^{'}} (0) = 3

y^{''}-7y^'-8y=0,y(0)=2,y^'(0)=7

y^{^{''}} - 7 y^{^{'}} - 8 y = 0, y (0) = 2, y^{^{'}} (0) = 7

Frequently Asked Questions (FAQ)

What is the solution for 4y^{''}+12y^'+9y=0,y(0)=1,y^'(0)=-4 ?
The solution for 4y^{''}+12y^'+9y=0,y(0)=1,y^'(0)=-4 is y=e^{-(3t)/2}-(5e^{-(3t)/2}t)/(2)