- #1
matematikawan
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We were trying to solve the problem 28.32 on page 289 of the Schaum's Series Differential Equations by Richard Bronson and Gabriel Costa. The DE is
[tex]4 x^2 y'' + (4 x + 2 x^2) y' + (3 x - 1) y = 0[/tex]
We use the Frobenius method to solve this equation since x=0 is a regular singular point. The difference in the indicial roots is an integer, i.e. [tex]\frac{1}{2} - \frac{-1}{2} = 1[/tex].
We suspect that the answer given in the book is incorrect since the expression for [tex]y_{2}(x)[/tex] does not contain a term like [tex]y_{1}ln(x)[/tex]. Since then we are searching for the correct answer to the problem.
Method 1
mail@riemann.physmath.fundp.ac.be sent me the following convode solution (if I simplified correctly)
[tex]y=\frac{cte}{8} ( \sqrt{x} \exp{(-\frac{x}{2})} ei(\frac{x}{2}) - \frac{2}{\sqrt{x}})\ + arbcomplex(1) \sqrt{x} \exp{(-\frac{x}{2})} [/tex]
Some explanation are in French language which I do not understand. I presume that term cte stand for constant , arbcomplex(1) is an arbitrary complex constant and ei(x) is an exponential integral (not sure of the exact definition). Anybody familiar with convode ?
Method 2
I try Mathematica and obtained the following
[tex]y[x] = A \sqrt{x}\exp{(-\frac{x}{2})} + B \sqrt{x}\exp{(-\frac{x}{2})} Gamma(-1,-\frac{x}{2}) [/tex].
But not so sure about the function Gamma(x,y).
Method 3
We use the method suggested in that Book. We obtained the first fundamental solution as
[tex]y_{1}(x)= \sqrt(x) (1 - \frac{x}{2} + \frac{x^2}{8} - \frac{x^3}{48}+...[/tex]
which is consistent with one of the solution given by the above softwares [tex]y_{1} = \sqrt{x} \exp{-\frac{x}{2}}[/tex].
To obtain the second fundamental solution we write
[tex]y(x)= a_{0} x^r (1 - \frac{x}{2r+1} + \frac{x^2}{(2r+1)(2r+3)} - \frac{x^3}{(2r+1)(2r+3)(2r+5)}+... [/tex].
Multiply by (2r + 1) and differentiate wrt r and substitute [tex]r_{2}=\frac{-1}{2} [/tex] we obtain
[tex]2 y_{2}(x)=-a_{0} \sqrt{x} (1 - \frac{x}{2} + \frac{x^2}{8} - ...)[/tex] [tex]+\frac{2a_{0}}{ \sqrt(x)} (1 - \frac{x^2}{4} + \frac{3x^3}{32} - ...)[/tex]
Do we work correctly ?
Method 4
We use the Lagrange Reduction of Order to obtain the second fundamental solution
[tex]y_{2} = u(x) y_{1} \ \ \ \mbox{where} \ \ \ u'(x) = x^{-2} \exp{(\frac{x}{2}}) [/tex]
Integrate
[tex]u(x)= - \frac{1}{x} + \frac{ln(x)}{2} +\frac{x}{8}+\frac{x^2}{96}+... [/tex].
Then
[tex]y_{2}(x)=\frac{1}{2} y_{1} ln(x) +\sqrt{x} \exp{(-\frac{x}{2}}) (- \frac{1}{x} + +\frac{x}{8}+\frac{x^2}{96}+...)[/tex].
Method 5
Use [tex]y_{2}(x)=d_{-1}y_{1} ln(x) + x^{r} \Sigma d_{n} x^n [/tex].
But we haven't try yet this method.
My question: Are the second fundamental solution obtain from methods 1 - 5 are all equal / equivalent ? I'm quite worry about the result obtained from method 4.
[tex]4 x^2 y'' + (4 x + 2 x^2) y' + (3 x - 1) y = 0[/tex]
We use the Frobenius method to solve this equation since x=0 is a regular singular point. The difference in the indicial roots is an integer, i.e. [tex]\frac{1}{2} - \frac{-1}{2} = 1[/tex].
We suspect that the answer given in the book is incorrect since the expression for [tex]y_{2}(x)[/tex] does not contain a term like [tex]y_{1}ln(x)[/tex]. Since then we are searching for the correct answer to the problem.
Method 1
mail@riemann.physmath.fundp.ac.be sent me the following convode solution (if I simplified correctly)
[tex]y=\frac{cte}{8} ( \sqrt{x} \exp{(-\frac{x}{2})} ei(\frac{x}{2}) - \frac{2}{\sqrt{x}})\ + arbcomplex(1) \sqrt{x} \exp{(-\frac{x}{2})} [/tex]
Some explanation are in French language which I do not understand. I presume that term cte stand for constant , arbcomplex(1) is an arbitrary complex constant and ei(x) is an exponential integral (not sure of the exact definition). Anybody familiar with convode ?
Method 2
I try Mathematica and obtained the following
[tex]y[x] = A \sqrt{x}\exp{(-\frac{x}{2})} + B \sqrt{x}\exp{(-\frac{x}{2})} Gamma(-1,-\frac{x}{2}) [/tex].
But not so sure about the function Gamma(x,y).
Method 3
We use the method suggested in that Book. We obtained the first fundamental solution as
[tex]y_{1}(x)= \sqrt(x) (1 - \frac{x}{2} + \frac{x^2}{8} - \frac{x^3}{48}+...[/tex]
which is consistent with one of the solution given by the above softwares [tex]y_{1} = \sqrt{x} \exp{-\frac{x}{2}}[/tex].
To obtain the second fundamental solution we write
[tex]y(x)= a_{0} x^r (1 - \frac{x}{2r+1} + \frac{x^2}{(2r+1)(2r+3)} - \frac{x^3}{(2r+1)(2r+3)(2r+5)}+... [/tex].
Multiply by (2r + 1) and differentiate wrt r and substitute [tex]r_{2}=\frac{-1}{2} [/tex] we obtain
[tex]2 y_{2}(x)=-a_{0} \sqrt{x} (1 - \frac{x}{2} + \frac{x^2}{8} - ...)[/tex] [tex]+\frac{2a_{0}}{ \sqrt(x)} (1 - \frac{x^2}{4} + \frac{3x^3}{32} - ...)[/tex]
Do we work correctly ?
Method 4
We use the Lagrange Reduction of Order to obtain the second fundamental solution
[tex]y_{2} = u(x) y_{1} \ \ \ \mbox{where} \ \ \ u'(x) = x^{-2} \exp{(\frac{x}{2}}) [/tex]
Integrate
[tex]u(x)= - \frac{1}{x} + \frac{ln(x)}{2} +\frac{x}{8}+\frac{x^2}{96}+... [/tex].
Then
[tex]y_{2}(x)=\frac{1}{2} y_{1} ln(x) +\sqrt{x} \exp{(-\frac{x}{2}}) (- \frac{1}{x} + +\frac{x}{8}+\frac{x^2}{96}+...)[/tex].
Method 5
Use [tex]y_{2}(x)=d_{-1}y_{1} ln(x) + x^{r} \Sigma d_{n} x^n [/tex].
But we haven't try yet this method.
My question: Are the second fundamental solution obtain from methods 1 - 5 are all equal / equivalent ? I'm quite worry about the result obtained from method 4.