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

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MATH257 December 2011

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

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

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.

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

Hint 3
Compare the ODE with $P(x)y^{″}+Q(x)y^{\prime }+R(x)y=0.$ 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?

Hint 4

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

Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

Solution

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 $y=\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^{n+r}.$ We have that $y=\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^{n+r},y^{\prime }=\sum _{n=0}^{\mathrm{\infty }}{a}_n(n+r)x^{n+r-1},y^{″}=\sum _{n=0}^{\mathrm{\infty }}{a}_n(n+r)(n+r-1)x^{n+r-2}.$ (Please note that the both of the sums for $y^{\prime }$ and $y^{″}$ start at $n=0$.) Substitute these into the ODE $2x^2{y}^{″}+(3x+x^2)y^{\prime }-y=0,$ we get that $\sum _{n=0}^{\mathrm{\infty }}2a_n(n+r)(n+r-1)x^{n+r}+\sum _{n=0}^{\mathrm{\infty }}3a_n(n+r)x^{n+r}+\sum _{n=0}^{\mathrm{\infty }}{a}_n(n+r)x^{n+r+1}-\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^{n+r}=0.$ We should shift the index $n$ of $\sum _{n=0}^{\mathrm{\infty }}{a}_n(n+r)x^{n+r+1}$ so the order of $x$ matches with the other terms. To do so, we should replace every $n$ with $n-1$. Then $\sum _{n=0}^{\mathrm{\infty }}{a}_n(n+r)x^{n+r+1}=\sum _{n=1}^{\mathrm{\infty }}{a}_{n-1}(n+r-1)x^{n+r}.$ Then we have $\sum _{n=0}^{\mathrm{\infty }}2a_n(n+r)(n+r-1)x^{n+r}+\sum _{n=0}^{\mathrm{\infty }}3a_n(n+r)x^{n+r}+\sum _{n=1}^{\mathrm{\infty }}{a}_{n-1}(n+r-1)x^{n+r}-\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^{n+r}=0$ which, after separating the $n=0$ terms, can be rewritten as $[2a_{0}r(r-1)+3a_{0}r-a_0]x^r+\sum _{n=1}^{\mathrm{\infty }}[2a_n(n+r)(n+r-1)+3a_n(n+r)+a_{n-1}(n+r-1)-a_n]x^{n+r}=0.$ 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 $[2a_n(n+r)(n+r-1)+3a_n(n+r)+a_{n-1}(n+r-1)-a_n]=0\text{for each}n\ge 1.$ If we Isolate $a_n$, we get the recurrence relation $a_n=\frac{-(n+r-1)a_{n-1}}{2(n+r)(n+r-1)+3(n+r)-1}\text{for}n\ge 1.$ 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 $r=-1$, $y=\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^{n-1}=a_0{x}^{-1}+a_1+a_{2}x+\cdots$. Note that the above equation does not satisfy $\underset{x\to 0^{+}}{\lim }y(x)=0$ 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 $r=\frac{1}{2}$ , $y=\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^{n+\frac{1}{2}}=a_0{x}^{\frac{1}{2}}+a_1{x}^{\frac{3}{2}}+a_2{x}^{\frac{5}{2}}+\cdots$. Note that the above equation satisfies $\underset{x\to 0^{+}}{\lim }y(x)=0$. 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 $y(x)=a_0{x}^{\frac{1}{2}}+a_1{x}^{\frac{3}{2}}+a_2{x}^{\frac{5}{2}}+\cdots =a_0{x}^{\frac{1}{2}}-\frac{1}{10}{a}_0{x}^{\frac{3}{2}}+\frac{3}{280}{a}_0{x}^{\frac{5}{2}}+\cdots$

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