# solve the initial value problem y"+y'-2y=-4; y(0) = y'(0)=0 using Laplace transformsy"+y'-2y=-4; y(0) = y'(0)=0

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### 1 Answer

To solve the IVP `y''+y'-2y=-4` with `y(0)=y'(0)=0`, we need to use the following Laplace transforms:

`L(y'')=s^2Y-sy(0)-y'(0)=s^2Y`

`L(y')=sY-sy(0)=sY`

`L(y)=Y`

`L(1)=1/s`

Applying the Laplace transform to the IVP gives the algebraic equation

`s^2Y+sY-2Y=-4/s`

`(s^2+s-2)Y=-4/s`

`Y=-4/{s(s+2)(s-1)}`

Now the RHS needs to be separated using partial fractions and comparing coefficients

`Y=A/s+B/{s+2}+C/{s-1}`

Comparing coefficients gives `A=2`, `B=-2/3` and `C=-4/3`.

Finally, apply the Inverse Laplace transform to Y to get

`Y=2/s-2/{3(s+2)}-4/{3(s-1)}`

`y(t)=2-2/3 e^{-2t}-4/3 e^t`

**The solution to the IVP using Laplace transforms is `y(t)=2-2/3 e^{-2t}-4/3e^t`.**