Question

What is the solution to the Initial Value Problem (IVP) `y''=2e^(-x)` with the IVs `y(0)=1,y'(0)=0`?

Answer 1

`y(x)=2e^(-x)+2x-1`

Explanation:

`y''=2e^(-x)--(1)`

`y(0)=1---(2)`

`y'(0)=0---(3)`

integrating (1) once we get

`y'(x)=-2e^(-x)+C_1`

using the initial condition `(3)`

`0=-2e^0+C_1`

`0=-2+C_1=>C_1=2`

`:.y'(x)=-2e^(-x)+2`

integrating once more

`y(x)=2e^(-x)+2x+C_2`

using the initial condition `(2)`

`1=2e^0+2xx0+C_2`

`1=2+0+C_2`

`C_2=-1`

`:.y(x)=2e^(-x)+2x-1`

Answer 2

`f(x) = 2e^(-x) + 2x -1`

Explanation:

As this is an IVP (Initial Value Problem) we can use Laplace Transforms:.

We have:

`y''=2e^(-x)` with the IVs `y(0)=1,y'(0)=0`

If we take Laplace Transformations of both sides of the above equation then we get:

`ℒ \ {y''} = ℒ \ {2e^(-x) }`

Then using the known property of the LT:

`ℒ \ {y''} =s^2 F(s)−s f(0)−f'(0)`
`ℒ \ {e^(at)} =1/(s-a)`

Then we have:

`s^2 F(s)−s f(0)−f'(0) = 2/(s+1)`

`:. s^2 F(s)-s-0 = 2/(s+1)`
`:. s^2 F(s) = s+2/(s+1)`
`:. s^2 F(s) = (s(s+1)+2)/(s+1)`
`:. s^2 F(s) = (s^2+s+2)/(s+1)`

`:. F(s) = (s^2+s+2)/(s^2(s+1))`

Now we can use partial fraction to decompose this expression:

`(s^2+s+2)/(s^2(s+1)) -= A/s + B/s^2 + C/(s+1)`
`=> s^2+s+2-= As(s+1) + B(s+1) + Cs^2`

We can find the constant coefficient as follows:

Put `s=0 \ \ \ \ \ => 2=0+B+0 => B=2`
Put `s=-1 => 2=0+0+C => C=2`
Compare `Coef(s^2) => 1=A+C => A=-1`

Thus we have:

`F(s) = -1/s + 2/s^2 +2/(s+1)`

Now if we take Inverse Laplace Transformations we have:

`ℒ^(-1) \ {F(s) } = ℒ^(-1) \ {-1/s} + ℒ^(-1) \ {2/s^2} +ℒ^(-1) \ {2/(s+1)}`

Again using the known property of the LT (inverses):

`f(x) = -1 + 2x +2e^(-x)`

Hence the complete solution is:

`f(x) = 2e^(-x) + 2x -1`