Science:Math Exam Resources/Courses/MATH257/December 2011/Question 01

MATH257 December 2011

• Q1 • Q2 (a) • Q2 (b) • Q3 (a) • Q3 (b) • Q4 (a) • Q4 (b) • Q5 (a) • Q5 (b) •

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• December 2011 •

[hide] Question 01
For the ordinary differential equation $2x^{2}y''+(3x+x^{2})y'-y=0,\,$ find the first 3 terms of a non-zero series solution about $x=0$ satisfying $\lim _{x\to 0^{+}}y(x)=0$ .

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[show] Hint 1
This is a standard question in this course. If you have forgotten how to do this type of question, check your class notes or homework for a similar example.

[show] Hint 2
Is $x=0$ an ordinary point or a singular point?

[show] Hint 3
Compare the ODE with $P(x)y''+Q(x)y'+R(x)y=0.$ We see that $P(x)=2x^{2}$ . Since $P(x)=0$ when $x=0$ , $x=0$ is a singular point. Is $x=0$ a regular singular point or an irregular singular point?

[show] Hint 4
Again compare the ODE with $P(x)y''+Q(x)y'+R(x)y=0.$ We see that $P(x)=2x^{2}$ , $Q(x)=3x+x^{2}$ and $R(x)=-1$ . Now, $\lim _{x\to 0}x{\frac {Q(x)}{P(x)}}=\lim _{x\to 0}x{\frac {3x+x^{2}}{2x^{2}}}=\lim _{x\to 0}({\frac {3}{2}}+{\frac {1}{2}}x)\;\;{\text{is finite}}$ and $\lim _{x\to 0}x^{2}{\frac {R(x)}{P(x)}}=\lim _{x\to 0}x^{2}{\frac {-1}{2x^{2}}}=-{\frac {1}{2}}\;\;{\text{is finite}}.$ Therefore, $x=0$ is a regular singular point. Should we consider series solutions of the form $y(x)=\sum _{n=0}^{\infty }a_{n}x^{n}\;\;{\text{or}}\;\;y(x)=\sum _{n=0}^{\infty }a_{n}x^{n+r}?$

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[show] Solution
Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Following the comments in the hints, we see that $x=0$ is a regular singular point. Therefore, we should look for series solutions of the form $y=\sum _{n=0}^{\infty }a_{n}x^{n+r}.$ We have that $y=\sum _{n=0}^{\infty }a_{n}x^{n+r},\;y'=\sum _{\color {red}n=0}^{\infty }a_{n}(n+r)x^{n+r-1},\;y''=\sum _{\color {red}n=0}^{\infty }a_{n}(n+r)(n+r-1)x^{n+r-2}.$ (Please note that the both of the sums for $y'$ and $y''$ start at $n=0$ .) Substitute these into the ODE $2x^{2}y''+(3x+x^{2})y'-y=0,$ we get that $\sum _{n=0}^{\infty }2a_{n}(n+r)(n+r-1)x^{n+r}+\sum _{n=0}^{\infty }3a_{n}(n+r)x^{n+r}{\color {blue}+\sum _{n=0}^{\infty }a_{n}(n+r)x^{n+r+1}}-\sum _{n=0}^{\infty }a_{n}x^{n+r}=0.$ We should shift the index $n$ of ${\color {blue}\sum _{n=0}^{\infty }a_{n}(n+r)x^{n+r+1}}$ so the order of $x$ matches with the other terms. To do so, we should replace every $n$ with $n-1$ . Then ${\color {blue}\sum _{n=0}^{\infty }a_{n}(n+r)x^{n+r+1}}=\sum _{n={\color {red}1}}^{\infty }a_{n{\color {red}-1}}(n+r{\color {red}-1})x^{\color {red}n+r}.$ Then we have $\sum _{n=0}^{\infty }2a_{n}(n+r)(n+r-1)x^{n+r}+\sum _{n=0}^{\infty }3a_{n}(n+r)x^{n+r}{\color {red}+\sum _{n=1}^{\infty }a_{n-1}(n+r-1)x^{n+r}}-\sum _{n=0}^{\infty }a_{n}x^{n+r}=0$ which, after separating the $n=0$ terms, can be rewritten as $[2a_{0}r(r-1)+3a_{0}r-a_{0}]\;x^{r}+\sum _{n=1}^{\infty }\;{\color {OliveGreen}[2a_{n}(n+r)(n+r-1)+3a_{n}(n+r)+a_{n-1}(n+r-1)-a_{n}]}\;x^{n+r}=0.$ Look at the above equation, because the right hand side is 0, the coefficient of the $x^{r}$ term must be 0 as well. Hence, $[2a_{0}r(r-1)+3a_{0}r-a_{0}]=a_{0}(2r(r-1)+3r-1)=0.\,$ Now, if $a_{0}=0$ , then all the subsequent $a_{n}$ will be zero as well and we only get the trivial solution $y=0$ . (Why? You can verify that yourself after we finish the question. Set $a_{0}=0$ and you will see the other terms will be forced to be 0 by the recursion relation.) In other words, we must have $2r(r-1)+3r-1=2r^{2}+r-1=(2r-1)(r+1)=0\,$ giving us two roots $r={\frac {1}{2}}$ and $r=-1\,$ . (Note: In the above, $2r(r-1)+3r-1=0$ is known as the indicial equation.) By the same reason as $[2a_{0}r(r-1)+3a_{0}r-a_{0}]=0$ , we have that ${\color {OliveGreen}[2a_{n}(n+r)(n+r-1)+3a_{n}(n+r)+a_{n-1}(n+r-1)-a_{n}]}=0\;\;{\text{for each}}\;\;n\geq 1.\,$ If we Isolate $a_{n}$ , we get the recurrence relation $a_{n}={\frac {-(n+r-1)a_{n-1}}{2(n+r)(n+r-1)+3(n+r)-1}}\;\;{\text{for}}\;\;n\geq 1.$ After simplifying and factoring the expression in the denominator, the above expression can be rewritten as $a_{n}={\frac {-(n+r-1)a_{n-1}}{2(n+r)^{2}+(n+r)-1}}={\frac {-(n+r-1)a_{n-1}}{(2(n+r)-1)((n+r)+1)}}$ which may be easier to do calculations with. Now recall that from the indicial equation, we found $r={\frac {1}{2}}$ and $r=-1\,$ . In the case $r=-1\,$ , $y=\sum _{n=0}^{\infty }a_{n}x^{n-1}=a_{0}x^{-1}+a_{1}+a_{2}x+\cdots$ . Note that the above equation does not satisfy $\lim _{x\to 0^{+}}y(x)=0$ due to the $a_{0}x^{-1}$ term, so the expression corresponding to $r=-1$ is not what the question is asking for. In the case $r={\frac {1}{2}}$ , $y=\sum _{n=0}^{\infty }a_{n}x^{n+{\frac {1}{2}}}=a_{0}x^{\frac {1}{2}}+a_{1}x^{\frac {3}{2}}+a_{2}x^{\frac {5}{2}}+\cdots$ . Note that the above equation satisfies $\lim _{x\to 0^{+}}y(x)=0$ . In other words, this is the one the question is asking for. It remains to find $a_{1}$ and $a_{2}$ in terms of $a_{0}$ . To find $a_{1}$ , substitue $n=1\,$ and $r={\frac {1}{2}}$ into the recurrence relation. We get that $a_{1}={\frac {-(1+{\frac {1}{2}}-1)a_{0}}{(2(1+{\frac {1}{2}})-1)(1+{\frac {1}{2}}+1)}}={\frac {-1}{10}}a_{0}.$ To find $a_{2}$ , substitue $n=2\,$ and $r={\frac {1}{2}}$ into the recurrence relation. We get that $a_{2}={\frac {-(2+{\frac {1}{2}}-1)a_{1}}{(2(2+{\frac {1}{2}})-1)(2+{\frac {1}{2}}+1)}}={\frac {-3}{28}}a_{1}={\frac {-3}{28}}({\frac {-1}{10}}a_{0})={\frac {3}{280}}a_{0}.$ Therefore, we have $y(x)=a_{0}x^{\frac {1}{2}}+a_{1}x^{\frac {3}{2}}+a_{2}x^{\frac {5}{2}}+\cdots =a_{0}x^{\frac {1}{2}}-{\frac {1}{10}}a_{0}x^{\frac {3}{2}}+{\frac {3}{280}}a_{0}x^{\frac {5}{2}}+\cdots$