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#
given that
1 Answer
# psi(1) = 1+e^(-2) #
Explanation:
The given question is in error, as
# 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) => 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 = v(x) e^(-2x) #
Differentiating wrt
# 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' - 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,
# d/dx{ (e^(-2x))/(2x+1) V } = 0 #
Which we can integrate to get:
# (e^(-2x))/(2x+1) V = A #
And, restoring the
# 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
# 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) = 1+e^(-2) #