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

#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#

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 #