Science:Math Exam Resources/Courses/MATH103/April 2015/Question 09

MATH103 April 2015

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Question 09
Suppose the power series $y=\sum _{n=0}^{\infty }a_{n}x^{n}$ solves the differential equation ${\frac {dy}{dx}}+xy=2$ with the initial condition $y(0)=-1$. Determine $a_{0}$, $a_{1}$, $a_{2}$, and $a_{3}$.

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 hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Substitute the given $y$ into the equation, and note that the index of summation can be shifted as the following rule: $\sum _{n=0}^{\infty }b_{n}=\sum _{n=1}^{\infty }b_{n-1}.$

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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. Since $y=\sum _{n=0}^{\infty }a_{n}x^{n}=a_{0}+\sum _{n=1}^{\infty }a_{n}x^{n}$, we can easily get $a_{0}=y(0)=-1$ from the initial condition. To get $a_{1}$, $a_{2}$, and $a_{3}$, plug $y=\sum _{n=0}^{\infty }a_{n}x^{n}$ into the equation. Then, using Hint 1, we have ${\frac {dy}{dx}}={\frac {d}{dx}}\left(\sum _{n=0}^{\infty }a_{n}x^{n}\right)=\sum _{n=0}^{\infty }{\frac {d}{dx}}(a_{n}x^{n})={\frac {d}{dx}}(a_{0})+\sum _{n=1}^{\infty }{\frac {d}{dx}}(a_{n}x^{n})=\sum _{n=1}^{\infty }a_{n}nx^{n-1}=\sum _{n=0}^{\infty }a_{n+1}(n+1)x^{n},$ and $xy=x\left(\sum _{n=0}^{\infty }a_{n}x^{n}\right)=\sum _{n=0}^{\infty }a_{n}x^{n+1}=\sum _{n=1}^{\infty }a_{n-1}x^{n},$ so that the equation can be written as ${\begin{aligned}2={\frac {dy}{dx}}+xy&=\sum _{n=1}^{\infty }a_{n}nx^{n-1}+\sum _{n=0}^{\infty }a_{n}x^{n+1}\qquad {\text{ we should match the exponents of x, so we re-index}}\\&=\sum _{n=0}^{\infty }a_{n+1}(n+1)x^{n}+\sum _{n=1}^{\infty }a_{n-1}x^{n}\\&=a_{1}+\sum _{n=1}^{\infty }a_{n+1}(n+1)x^{n}+\sum _{n=1}^{\infty }a_{n-1}x^{n}\\&=a_{1}+\sum _{n=1}^{\infty }(a_{n+1}(n+1)+a_{n-1})x^{n}\\&=a_{1}+(2a_{2}+a_{0})x+(3a_{3}+a_{1})x^{2}+\cdots .\end{aligned}}$ Now comparing each coefficient of $x^{n}$ on both side, we get ${\begin{aligned}a_{1}&=2\\2a_{2}+a_{0}&=0\\3a_{3}+a_{1}&=0.\end{aligned}}$ Using $a_{0}=-1$, the second equation implies that $a_{2}={\frac {1}{2}}$. On the other hand, plugging $a_{1}=2$ into the third equation, we obtain $a_{3}=-{\frac {2}{3}}$. Therefore, the answer is $\color {blue}a_{0}=-1,\quad a_{1}=2,\quad a_{2}={\frac {1}{2}},\quad a_{3}=-{\frac {2}{3}}$.