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Popular >

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

Solution

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

Solution

y = (\frac{3}{2} - \frac{1}{10}) e^{2 t} - \frac{3}{2} e^{t} + \frac{1}{10} e^{- 3 t}

Solution steps

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

Solve linear ODE:

y = (\frac{3}{2} - \frac{1}{10}) e^{2 t} - \frac{3}{2} e^{t} + \frac{1}{10} e^{- 3 t}

y = (\frac{3}{2} - \frac{1}{10}) e^{2 t} - \frac{3}{2} e^{t} + \frac{1}{10} e^{- 3 t}

Graph

Popular Examples

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

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

y^{''}+4y=e^{-4t},y(0)=2

y^{^{''}} + 4 y = e^{- 4 t}, y (0) = 2

y^{''}+25y=40cos(5t),y(0)=y

y^{^{''}} + 25 y = 40 cos (5 t), y (0) = y

y^{''}-4y^'+8y=e^{2x}+sin(2x)

y^{^{''}} - 4 y^{^{'}} + 8 y = e^{2 x} + sin (2 x)

y^{''}+25y=sec^3(5t)

y^{^{''}} + 25 y = sec^{3} (5 t)

Frequently Asked Questions (FAQ)

What is the solution for y^{''}-3y^'+2y=2e^{-3t},y(0)=0,y^'(0)=1 ?
The solution for y^{''}-3y^'+2y=2e^{-3t},y(0)=0,y^'(0)=1 is y=(3/2-1/10)e^{2t}-3/2 e^t+1/10 e^{-3t}