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Popular >

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

Solution

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

Solution

y = c_{1} + c_{2} e^{- \frac{t}{4}}

Solution steps

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

Solve linear ODE:

y = c_{1} + c_{2} e^{- \frac{t}{4}}

y = c_{1} + c_{2} e^{- \frac{t}{4}}

Graph

Popular Examples

y^{''''}+2y^{'''}+4y^{''}+8y^'=0

y^{^{''''}} + 2 y^{^{'''}} + 4 y^{^{''}} + 8 y^{^{'}} = 0

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

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

2(d^2y)/(dx^2)+8(dy)/(dx)-y=0

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

y^{''}+10y^'+8/3 y=0

y^{^{''}} + 10 y^{^{'}} + \frac{8}{3} y = 0

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

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

Frequently Asked Questions (FAQ)

What is the solution for y^'+4y^{''}=0 ?
The solution for y^'+4y^{''}=0 is y=c_{1}+c_{2}e^{-t/4}