Science:Math Exam Resources/Courses/MATH103/April 2017/Question 08 (b)

MATH103 April 2017

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Question 08 (b)
Suppose the Taylor series $y=\sum _{n=0}^{\infty }a_{n}x^{n}$ solves the differential equation ${\frac {dy}{dx}}+2y=x^{2}$ with the initial condition $y(0)=4$. (b) Find the recursive relation $a_{n}=f(a_{n-1})$ for $n\geq 4$.

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
To find the recursive relation, collect all like terms after expressing the differential equation ${\frac {dy}{dx}}+2y=x^{2}$ entirely in terms of Taylor series.

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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. Like in part a), write the differential equation ${\frac {dy}{dx}}+2y=x^{2}$ entirely in terms of Taylor series to obtain $(2a_{0}+a_{1})+(2a_{1}+2a_{2})x+(2a_{2}+3a_{3})x^{2}+(2a_{3}+4a_{4})x^{3}+\dots =0+0x+x^{2}+0x^{3}+0x^{4}+\dots$ Collecting like terms, we obtain $(2a_{0}+a_{1})+(2a_{1}+2a_{2})x+(2a_{2}+3a_{3}-1)x^{2}+(2a_{3}+4a_{4})x^{3}+\dots =0.$ This can be re-written as $(2a_{0}+a_{1})+(2a_{1}+2a_{2})x+(2a_{2}+3a_{3}-1)x^{2}+\sum _{n=4}^{\infty }(2a_{n-1}+na_{n})x^{n-1}=0=\sum _{n=4}^{\infty }0x^{n-1}.$ Hence, for all $n\geq 4$, $2a_{n-1}+na_{n}=0.$ In other words, the following holds for all $n\geq 4$, $\color {blue}a_{n}={\frac {-2}{n}}a_{n-1}$