Science:Math Exam Resources/Courses/MATH103/April 2011/Question 02 (c)

MATH103 April 2011

  •  Q1 (a)  •  Q1 (b)  •  Q1 (c)  •  Q2 (a)  •  Q2 (b)  •  Q2 (c)  •  Q2 (d)  •  Q3 (a)  •  Q3 (b)  •  Q4 (a)  •  Q4 (b)  •  Q5 (a)  •  Q5 (b)  •  Q5 (c)  •  Q5 (d)  •  Q6 (a)  •  Q6 (b)  •  Q7 (a)  •  Q7 (b)  •  Q7 (c)  •  Q8  •  

   Other MATH103 Exams

  •  April 2006  •  April 2005  •  April 2011  •  April 2010  •  April 2009  •  April 2012  •  April 2013  •  April 2014  •  April 2015  •  April 2016  •  April 2017  •  

Question 02 (c)
Consider the differential equation ${\frac {dy}{dx}}=y-x.$ Use a Taylor series expansion to find the solution $y(x)$ for the initial value $y(0)=1$.

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
Write $y=y(x)$ as a Taylor series about $x=0,$ $y=\sum _{n=0}^{\infty }a_{n}x^{n},$ where the coefficients $a_{n}$ are to be determined.

Hint 2
Use the initial condition and the differential equation to solve for the coefficients in the Taylor Series.

Hint 3
From the initial condition, $y(0)=1,$ what can you conclude about $a_{0}$?

Hint 4
Differentiate the series term by term and compare with the series for $y-x$.

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

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Assume that the solution $y=y(x)$ has a Taylor series expansion about $x=0,$ $y=a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}+\dots =\sum _{n=0}^{\infty }a_{n}x^{n}.$ We'll use the initial condition and the differential equation to solve for the coeffients. From the initial condition, $1=y(0)=a_{0}+a_{1}(0)+a_{2}(0)^{2}+a_{3}(0)^{3}+\dots =a_{0}.$ Let's start with the left hand side $dy/dx$ of the differential equation $dy/dx=y-x$. Differentiating the series for $y$ term by term yields ${\begin{aligned}{\frac {dy}{dx}}&=0+a_{1}+2a_{2}x+3a_{3}x^{2}+4a_{4}x^{3}+\dots \\&=\sum _{n=1}^{\infty }a_{n}nx^{n-1}\\&=\sum _{k=0}^{\infty }a_{k+1}(k+1)x^{k}\\&=\sum _{n=0}^{\infty }a_{n+1}(n+1)x^{n}\\\end{aligned}}$ Here we substituted the dummy variable $n=k+1$, (i.e. $k=n-1$) in the third step, and then renamed $k$ as $n$. (This substitution of dummy variable was done purely for <<cosmetic reasons>> to be able to compare the two Taylor series of the LHS and the RHS of the differential equation more easily. In general it helps with both series start with $n=0$.) The right hand side $y-x$ of the differential equation $dy/dx=y-x$ gives ${\begin{aligned}y-x&=a_{0}+(a_{1}-1)x+a_{2}x^{2}+a_{3}x^{3}+\dots \\&=a_{0}+(a_{1}-1)x+\sum _{n=2}^{\infty }a_{n}x^{n}.\end{aligned}}$ Now we can equate the left hand side with the right hand side of the differential equation to get Failed to parse (unknown function "\begin{align}"): {\displaystyle \begin{align} \frac{dy}{dx} &= y-x \\ {\color{green} a_1} + {\color{red} 2a_2x} + {\color{blue} 3a_3x^2} + {\color{magenta} 4a_4x^3} + \dots &= {\color{green} a_0} + {\color{red} (a_1-1)x} + {\color{blue} a_2x^2} + {\color{magenta} a_3x^3} + \dots \\ {\color{green} a_1} + {\color{red} 2a_2x} + \sum_{n=2}^\infty a_{n+1} (n+1) x^{n} &= {\color{green} a_0} + {\color{red} (a_1-1)x} + \sum_{n=2}^\infty a_n x^n. \end{align} } From the differential equation, we can equate each of the coefficients of the above series. For $n=0,$ (i.e. the coefficients in from of x0) we have ${\color {green}a_{1}}={\color {green}a_{0}}=1.$ For $n=1,$ (i.e. the coefficients in from of x1) ${\color {red}2a_{2}}={\color {red}a_{1}-1}=0,$ hence, $a_{2}=0.$ For all $n\geq 2,$ $a_{n+1}(n+1)=a_{n}$. That is, $a_{n+1}={\frac {a_{n}}{n+1}}.$ For example, ${\color {blue}a_{3}}={\color {blue}a_{2}/3}=0.$ Similarly, ${\color {magenta}a_{4}}={\color {magenta}a_{3}/4}=0.$ We can see that for all $n\geq 3,$ $a_{n}=0.$ Hence, the solution is $\displaystyle y=1+x.$ Note: We can easily check that this is indeed a solution to the initial value problem: $y(0)=1+0=1$ and $dy/dx=1=y-x$.

Click here for similar questions

MER QGH flag, MER QGQ flag, MER QGS flag, MER RT flag, MER Tag Initial value problem, MER Tag Taylor series, Pages using DynamicPageList3 parser function, Pages using DynamicPageList3 parser tag, Pages with math errors, Pages with math render errors