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Popular >

y^{''}+3y^'+2y=e^t,y(0)=1,y^'(0)=-6

Solution

y^{^{''}} + 3 y^{^{'}} + 2 y = e^{t}, y (0) = 1, y^{^{'}} (0) = - 6

Solution

y = (- \frac{16}{3} + \frac{5}{6}) e^{- t} + \frac{16}{3} e^{- 2 t} + \frac{1}{6} e^{t}

Solution steps

y^{''} (t) + 3 y^{'} (t) + 2 y = e^{t}

Solve linear ODE:

y = (- \frac{16}{3} + \frac{5}{6}) e^{- t} + \frac{16}{3} e^{- 2 t} + \frac{1}{6} e^{t}

y = (- \frac{16}{3} + \frac{5}{6}) e^{- t} + \frac{16}{3} e^{- 2 t} + \frac{1}{6} e^{t}

Popular Examples

y^{''}+2y^'+4y=111e^{2t}cos(3t)

y^{^{''}} + 2 y^{^{'}} + 4 y = 111 e^{2 t} cos (3 t)

y^{''}+4y= 1/(cos(2x))

y^{^{''}} + 4 y = \frac{1}{cos ( 2 x )}

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

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

(d^2y)/(dx^2)-2(dy)/(dx)=x+sin(x)

\frac{d ^{2} y}{d x ^{2}} - 2 \frac{d y}{d x} = x + sin (x)

y^{''}-4y^'+4y=3e^{2x}

y^{^{''}} - 4 y^{^{'}} + 4 y = 3 e^{2 x}

Frequently Asked Questions (FAQ)

What is the solution for y^{''}+3y^'+2y=e^t,y(0)=1,y^'(0)=-6 ?
The solution for y^{''}+3y^'+2y=e^t,y(0)=1,y^'(0)=-6 is y=(-16/3+5/6)e^{-t}+16/3 e^{-2t}+1/6 e^t