Find the value ϕ(1) for a solution ϕ(x) to the initial value problem: #(1+2x)y′′+4xy′−4y=0; y(0)=1;y′(0)=−1# ?

given that #e^(2x)# is a solution to #(1+2x)y′′+4xy′−4y=0#

1 Answer
Steve M
Feb 7, 2018

# psi(1) = 1+e^(-2) #

Explanation:

The given question is in error, as #y=e^(2x)# is not a solution of the given DE. We can readily show this to be the case:

# y=e^(2x) => y' = 2e^(2x) \ \ , y'' = 4e^(2x) #

Then:

# (1+2x)y''+4xy'−4y = (1+2x)4e^(2x)+4x2e^(2x)−4e^(2x) #
# " " = 4e^(2x)+8xe^(2x) +8xe^(2x)−4e^(2x) #
# " " = 16xe^(2x) #
# " " != 0 #

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Having shown this, we can however verify that, #y=e^(-2x)# is a solution, viz:

# y=e^(-2x) => y' = -2e^(-2x) \ \ , y'' = 4e^(-2x) #

Then:

# (1+2x)y''+4xy'−4y = (1+2x)4e^(-2x)-4x2e^(-2x)−4e^(2x) #
# " " = 4e^(2x)+8xe^(2x) -8xe^(2x)−4e^(2x) #
# " " = 0 #

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Using this major, and fundamental amendment to the specified question, we can proceed as follows:

We have:

# (1+2x)y''+4xy'−4y = 0 #

We are given that a solution is #y=e^(-2x)#, Knowing this we assume we can find another solution of the form:

# y = v(x) e^(-2x) #

Differentiating wrt #x# we get:

# y' \ = v'e^(-2x) -2ve^(-2x)#
# y'' = v''e^(-2x) - 4v'e^(-2x) +4ve^(-2x)#

Then substituting into the DE, we get:

# (1+2x){v''e^(-2x) - 4v'e^(-2x) +4ve^(-2x)}+4x{v'e^(-2x) -2ve^(-2x)}−4ve^(-2x) = 0 #

# :. v'' + 2xv'' - 4v' - 4xv' = 0 #

# :. (1+2x)v'' - 4(1+x)v' = 0 #

If we write #V=v'#, then we get:

# V' - 4(1+x)/(1+2x)V = 0 # ..... [A]

Which is a First Order, ODE solvable using an integrating factor, I#I, where:

# I =exp( int \ - 4(1+x)/(1+2x) \ dx )#
# \ \ = (e^(-2x))/(2x+1) #

Multiplying the DE [A] by the IF, #I#, we get:

# d/dx{ (e^(-2x))/(2x+1) V } = 0 #

Which we can integrate to get:

# (e^(-2x))/(2x+1) V = A #

And, restoring the #V# suibstitution we have:

# v' = A(2x+1)e^(2x) #

Then after integration by parts, we can integrate again to get:

# v = Axe^(2x) + B #

And, now restoring the #v# substitution, we gte:

# y = ve^(-2x) #
# \ \ = Ax + Be^(-2x) #

And given this if we differentiate, we find that:

# y'(x) = A -2Be^(-2x) #

We can now use the initial conditions:

# y(0)=1 => B = 1 #
# y'(0)=-1 => A -2 = -1 => A=1#

Leading to the full solution:

# y(x) = x + e^(-2x) #

Then if we require, #psi(1)#, where, #psi(x)=y(x)#, then:

# psi(1) = 1+e^(-2) #

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