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

Jun 24, 2017

$y \left(x\right) = 2 {e}^{- x} + 2 x - 1$

#### Explanation:

$y ' ' = 2 {e}^{- x} - - \left(1\right)$

$y \left(0\right) = 1 - - - \left(2\right)$

$y ' \left(0\right) = 0 - - - \left(3\right)$

integrating (1) once we get

$y ' \left(x\right) = - 2 {e}^{- x} + {C}_{1}$

using the initial condition $\left(3\right)$

$0 = - 2 {e}^{0} + {C}_{1}$

$0 = - 2 + {C}_{1} \implies {C}_{1} = 2$

$\therefore y ' \left(x\right) = - 2 {e}^{- x} + 2$

integrating once more

$y \left(x\right) = 2 {e}^{- x} + 2 x + {C}_{2}$

using the initial condition $\left(2\right)$

$1 = 2 {e}^{0} + 2 \times 0 + {C}_{2}$

$1 = 2 + 0 + {C}_{2}$

${C}_{2} = - 1$

$\therefore y \left(x\right) = 2 {e}^{- x} + 2 x - 1$

Jun 24, 2017

$f \left(x\right) = 2 {e}^{- x} + 2 x - 1$

#### Explanation:

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

We have:

$y ' ' = 2 {e}^{- x}$ with the IVs $y \left(0\right) = 1 , y ' \left(0\right) = 0$

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)

$\therefore {s}^{2} F \left(s\right) - s - 0 = \frac{2}{s + 1}$
$\therefore {s}^{2} F \left(s\right) = s + \frac{2}{s + 1}$
$\therefore {s}^{2} F \left(s\right) = \frac{s \left(s + 1\right) + 2}{s + 1}$
$\therefore {s}^{2} F \left(s\right) = \frac{{s}^{2} + s + 2}{s + 1}$

$\therefore F \left(s\right) = \frac{{s}^{2} + s + 2}{{s}^{2} \left(s + 1\right)}$

Now we can use partial fraction to decompose this expression:

$\frac{{s}^{2} + s + 2}{{s}^{2} \left(s + 1\right)} \equiv \frac{A}{s} + \frac{B}{s} ^ 2 + \frac{C}{s + 1}$
$\implies {s}^{2} + s + 2 \equiv A s \left(s + 1\right) + B \left(s + 1\right) + C {s}^{2}$

We can find the constant coefficient as follows:

Put $s = 0 \setminus \setminus \setminus \setminus \setminus \implies 2 = 0 + B + 0 \implies B = 2$
Put $s = - 1 \implies 2 = 0 + 0 + C \implies C = 2$
Compare $C o e f \left({s}^{2}\right) \implies 1 = A + C \implies A = - 1$

Thus we have:

$F \left(s\right) = - \frac{1}{s} + \frac{2}{s} ^ 2 + \frac{2}{s + 1}$

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):

$f \left(x\right) = - 1 + 2 x + 2 {e}^{- x}$

Hence the complete solution is:

$f \left(x\right) = 2 {e}^{- x} + 2 x - 1$