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
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)