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

MATH103 April 2017

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Question 08 (a)
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 $y(0)=4$. (a) Calculate the first four coefficients $a_{0},a_{1},a_{2}$ and $a_{3}$.

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Hint
Express the differential equation entirely in terms of Taylor series. Then collect like terms.

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Solution 1

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 we have the power rule for differentiation, we can differentiate any Taylor series to obtain another Taylor series. With $y=a_{0}+a_{1}x+a_{2}x^{2}+\dots$, we calculate ${\frac {dy}{dx}}=a_{1}+2a_{2}x+3a_{3}x^{3}+4a_{4}x^{3}+\dots$ So ${\frac {dy}{dx}}+2y=(a_{1}+2a_{2}x+3a_{3}x^{2}+4a_{4}x^{3}+\dots )+(2a_{0}+2a_{1}x+2a_{2}x^{2}+2a_{3}x^{3}+\dots )$ $=(2a_{0}+a_{1})+(2a_{1}+2a_{2})x+(2a_{2}+3a_{3})x^{2}+(2a_{3}+4a_{4})x^{3}+\dots$ Because $x^{2}=0+0x+x^{2}+0x^{3}+0x^{4}+\dots ,$ ${\frac {dy}{dx}}+2y=x^{2}$ becomes $(2a_{0}+a_{1})+(2a_{1}+2a_{2})x+(2a_{2}+3a_{3})x^{2}+(2a_{3}+4a_{4})x^{3}+\dots =0+0x+x^{2}+0x^{3}+0x^{4}+\dots$ Collecting like terms for the first three monomials on each side now gives $2a_{0}+a_{1}=0,2a_{1}+2a_{2}=0,{\text{ and }}2a_{2}+3a_{3}=1$ The first value $a_{0}$ can be determined from the initial condition $y(0)=4$ because $y(0)=\sum _{n=0}^{\infty }a_{n}0^{n}=a_{0}+a_{1}0+a_{2}0^{2}+a_{3}0^{3}+\dots =a_{0}.$ It follows that $a_{0}=y(0)=4$. Using the above three equations, one can then calculate the following as the answer: $\color {blue}a_{0}=4,a_{1}=-8,a_{2}=8,{\text{ and }}a_{3}=-5$

Solution 2

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. We know that the coefficients of the Taylor series are given by $a_{n}={\frac {y^{(n)}(0)}{n!}}$ and that the initial condition is $y(0)={\color {MediumVioletRed}4}.$ Since $y$ solves the differential equation $y'+2y=x^{2}$, or equivalently, $y'=x^{2}-2y$, we must have $y'(0)=0^{2}-2y(0)=0-2\cdot {\color {MediumVioletRed}4}={\color {Chocolate}-8}.$ By differentiating the equation for $y'$ (with respect to $x$), we obtain $y''=2x-2y'$, which implies that $y''(0)=2\cdot 0-2y'(0)=0-2\cdot {\color {Chocolate}-8}={\color {DarkGreen}16}.$ Differentiating the equation once more, we obtain $y'''=2-2y''$, so $y'''(0)=2-2y''(0)=2-2\cdot {\color {DarkGreen}16}=-30.$ We conclude that $a_{0}={\frac {y(0)}{0!}}=\color {blue}4$, $a_{1}={\frac {y'(0)}{1!}}=\color {blue}-8$, $a_{2}={\frac {y''(0)}{2!}}=\color {blue}8$, and $a_{3}={\frac {y'''(0)}{3!}}=\color {blue}-5$.