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

MATH103 April 2011

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[hide] 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.

[show] 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.

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

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

[show] 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.

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[show] Solution
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 $dy/dx=1=y-x$ .