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Popular Calculus >

y^{''}+2y^'-y=t^{-1}*e^t

Solution

y^{^{''}} + 2 y^{^{'}} - y = t^{- 1} \cdot e^{t}

Solution

y = c_{1} e^{(- 1 + 2) t} + c_{2} e^{(- 1 - 2) t} + e^{- t + 2 t} \cdot \int \frac{2 e ^{- 2 t + 2 t}}{4 t} d t + e^{- t - 2 t} \cdot \int - \frac{2 e ^{2 t + 2 t}}{4 t} d t

Solution steps

y^{''} (t) + 2 y^{'} (t) - y = t^{- 1} \cdot e^{t}

Solve linear ODE:

y = c_{1} e^{(- 1 + 2) t} + c_{2} e^{(- 1 - 2) t} + e^{- t + 2 t} \cdot \int \frac{2 e ^{- 2 t + 2 t}}{4 t} d t + e^{- t - 2 t} \cdot \int - \frac{2 e ^{2 t + 2 t}}{4 t} d t

y = c_{1} e^{(- 1 + 2) t} + c_{2} e^{(- 1 - 2) t} + e^{- t + 2 t} \cdot \int \frac{2 e ^{- 2 t + 2 t}}{4 t} d t + e^{- t - 2 t} \cdot \int - \frac{2 e ^{2 t + 2 t}}{4 t} d t

Popular Examples

y^{''}-2y^'+5y=2x+e^{-x}

x^{''}-x^'-12x=e^{4t},x(0)=1,x^'(0)=0

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

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

Frequently Asked Questions (FAQ)

What is the solution for y^{''}+2y^'-y=t^{-1}*e^t ?
The solution for y^{''}+2y^'-y=t^{-1}*e^t is