Series Solutions of Differential Equations

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1. Find a power series solution for the following differential equations.

   1. \({y}^{″}-y=0\)
   2. \(({x}^{2}-1){y}^{″}+6x{y}^{'}+4y=-4\)
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   1. Assume \(y(x)=\sum _{n=0}^{\infty }{a}_{n}{x}^{n}\) (step 1). Then, \({y}^{'}(x)=\sum _{n=1}^{\infty }n{a}_{n}{x}^{n-1}\) and \({y}^{″}(x)=\sum _{n=2}^{\infty }n(n-1){a}_{n}{x}^{n-2}\) (step 2). We want to find values for the coefficients \({a}_{n}\) such that
      \[\begin{array}{lll}{y}^{″}-y & = & 0 \\ \sum _{n=2}^{\infty }n(n-1){a}_{n}{x}^{n-2}-\sum _{n=0}^{\infty }{a}_{n}{x}^{n} & = & 0\ \text{(step 3).}\end{array}\]
      We want the indices on our sums to match so that we can express them using a single summation. That is, we want to rewrite the first summation so that it starts with \(n=0.\)
      To re-index the first term, replace n with \(n+2\) inside the sum, and change the lower summation limit to \(n=0.\) We get
      \[\sum _{n=2}^{\infty }n(n-1){a}_{n}{x}^{n-2}=\sum _{n=0}^{\infty }(n+2)(n+1){a}_{n+2}{x}^{n}.\]
      This gives
      \[\begin{array}{lll}\sum _{n=0}^{\infty }(n+2)(n+1){a}_{n+2}{x}^{n}-\sum _{n=0}^{\infty }{a}_{n}{x}^{n} & = & 0 \\ \sum _{n=0}^{\infty }[(n+2)(n+1){a}_{n+2}-{a}_{n}]{x}^{n} & = & 0\ \text{(step 4).}\end{array}\]
      Because power series expansions of functions are unique, this equation can be true only if the coefficients of each power of x are zero. So we have
      \[(n+2)(n+1){a}_{n+2}-{a}_{n}=0\ \text{for}\ n=0,1,2\text{,}\ldots \text{.}\]
      This recurrence relationship allows us to express each coefficient \({a}_{n}\) in terms of the coefficient two terms earlier. This yields one expression for even values of n and another expression for odd values of n. Looking first at the equations involving even values of n, we see that
      \[\begin{array}{lll}{a}_{2} & = & \frac{{a}_{0}}{2} \\ {a}_{4} & = & \frac{{a}_{2}}{4⋅3}=\frac{{a}_{0}}{4!} \\ {a}_{6} & = & \frac{{a}_{4}}{6⋅5}=\frac{{a}_{0}}{6!} \\ & \vdots \text{.} & \end{array}\]
      Thus, in general, when n is even, \({a}_{n}=\frac{{a}_{0}}{n!}\) (step 5).
      For the equations involving odd values of n, we see that
      \[\begin{array}{lll}{a}_{3} & = & \frac{{a}_{1}}{3⋅2}=\frac{{a}_{1}}{3!} \\ {a}_{5} & = & \frac{{a}_{3}}{5⋅4}=\frac{{a}_{1}}{5!} \\ {a}_{7} & = & \frac{{a}_{5}}{7⋅6}=\frac{{a}_{1}}{7!} \\ & \vdots \text{.} & \end{array}\]
      Therefore, in general, when n is odd, \({a}_{n}=\frac{{a}_{1}}{n!}\) (step 5 continued).
      Putting this together, we have
      \[\begin{array}{ll}y(x) & =\sum _{n=0}^{\infty }{a}_{n}{x}^{n} \\ & ={a}_{0}+{a}_{1}x+\frac{{a}_{0}}{2}{x}^{2}+\frac{{a}_{1}}{3!}{x}^{3}+\frac{{a}_{0}}{4!}{x}^{4}+\frac{{a}_{1}}{5!}{x}^{5}+\cdots \text{.}\end{array}\]
      Re-indexing the sums to account for the even and odd values of n separately, we obtain
      \[y(x)={a}_{0}\sum _{k=0}^{\infty }\frac{1}{(2k)!}{x}^{2k}+{a}_{1}\sum _{k=0}^{\infty }\frac{1}{(2k+1)!}{x}^{2k+1}\ \text{(step 6).}\]
      Analysis for part a.
      As expected for a second-order differential equation, this solution depends on two arbitrary constants. However, note that our differential equation is a constant-coefficient differential equation, yet the power series solution does not appear to have the familiar form (containing exponential functions) that we are used to seeing. Furthermore, since \(y(x)={c}_{1}{e}^{x}+{c}_{2}{e}^{\text{-}x}\) is the general solution to this equation, we must be able to write any solution in this form, and it is not clear whether the power series solution we just found can, in fact, be written in that form.
      Fortunately, after writing the power series representations of \({e}^{x}\) and \({e}^{\text{-}x},\) and doing some algebra, we find that if we choose
      \[{c}_{0}=\frac{({a}_{0}+{a}_{1})}{2},\ {c}_{1}=\frac{({a}_{0}-{a}_{1})}{2},\]
      we then have \({a}_{0}={c}_{0}+{c}_{1}\) and\({a}_{1}={c}_{0}-{c}_{1},\) and
      \[\begin{array}{ll}y(x) & ={a}_{0}+{a}_{1}x+\frac{{a}_{0}}{2}{x}^{2}+\frac{{a}_{1}}{3!}{x}^{3}+\frac{{a}_{0}}{4!}{x}^{4}+\frac{{a}_{1}}{5!}{x}^{5}+\cdots \\ & =({c}_{0}+{c}_{1})+({c}_{0}-{c}_{1})x+\frac{({c}_{0}+{c}_{1})}{2}{x}^{2}+\frac{({c}_{0}-{c}_{1})}{3!}{x}^{3}+\frac{({c}_{0}+{c}_{1})}{4!}{x}^{4}+\frac{({c}_{0}-{c}_{1})}{5!}{x}^{5}+\cdots \\ & ={c}_{0}\sum _{n=0}^{\infty }\frac{{x}^{n}}{n!}+{c}_{1}\sum _{n=0}^{\infty }\frac{{(\text{-}x)}^{n}}{n!} \\ & ={c}_{0}{e}^{x}+{c}_{1}{e}^{\text{-}x}.\end{array}\]
      So we have, in fact, found the same general solution. Note that this choice of \({c}_{1}\) and \({c}_{2}\) is not obvious. This is a case when we know what the answer should be, and have essentially “reverse-engineered” our choice of coefficients.
   2. Assume \(y(x)=\sum _{n=0}^{\infty }{a}_{n}{x}^{n}\) (step 1). Then, \({y}^{'}(x)=\sum _{n=1}^{\infty }n{a}_{n}{x}^{n-1}\) and \({y}^{″}(x)=\sum _{n=2}^{\infty }n(n-1){a}_{n}{x}^{n-2}\) (step 2). We want to find values for the coefficients \({a}_{n}\) such that
      \[\begin{array}{lll} \\ \\ ({x}^{2}-1){y}^{″}+6x{y}^{'}+4y & = & -4 \\ ({x}^{2}-1)\sum _{n=2}^{\infty }n(n-1){a}_{n}{x}^{n-2}+6x\sum _{n=1}^{\infty }n{a}_{n}{x}^{n-1}+4\sum _{n=0}^{\infty }{a}_{n}{x}^{n} & = & -4 \\ {x}^{2}\sum _{n=2}^{\infty }n(n-1){a}_{n}{x}^{n-2}-\sum _{n=2}^{\infty }n(n-1){a}_{n}{x}^{n-2}+6x\sum _{n=1}^{\infty }n{a}_{n}{x}^{n-1}+4\sum _{n=0}^{\infty }{a}_{n}{x}^{n} & = & -4.\end{array}\]
      Taking the external factors inside the summations, we get
      \[\sum _{n=2}^{\infty }n(n-1){a}_{n}{x}^{n}-\sum _{n=2}^{\infty }n(n-1){a}_{n}{x}^{n-2}+\sum _{n=1}^{\infty }6n{a}_{n}{x}^{n}+\sum _{n=0}^{\infty }4{a}_{n}{x}^{n}=-4\ \text{(step 3).}\]
      Now, in the first summation, we see that when \(n=0\) or \(n=1,\) the term evaluates to zero, so we can add these terms back into our sum to get
      \[\sum _{n=2}^{\infty }n(n-1){a}_{n}{x}^{n}=\sum _{n=0}^{\infty }n(n-1){a}_{n}{x}^{n}.\]
      Similarly, in the third term, we see that when \(n=0,\) the expression evaluates to zero, so we can add that term back in as well. We have
      \[\sum _{n=1}^{\infty }6n{a}_{n}{x}^{n}=\sum _{n=0}^{\infty }6n{a}_{n}{x}^{n}.\]
      Then, we need only shift the indices in our second term. We get
      \[\sum _{n=2}^{\infty }n(n-1){a}_{n}{x}^{n-2}=\sum _{n=0}^{\infty }(n+2)(n+1){a}_{n+2}{x}^{n}.\]
      Thus, we have
      \[\begin{array}{lll}\sum _{n=0}^{\infty }n(n-1){a}_{n}{x}^{n}-\sum _{n=0}^{\infty }(n+2)(n+1){a}_{n+2}{x}^{n}+\sum _{n=0}^{\infty }6n{a}_{n}{x}^{n}+\sum _{n=0}^{\infty }4{a}_{n}{x}^{n} & = & -4\ \text{(step 4).} \\ \sum _{n=0}^{\infty }[n(n-1){a}_{n}-(n+2)(n+1){a}_{n+2}+6n{a}_{n}+4{a}_{n}]{x}^{n} & = & -4 \\ \sum _{n=0}^{\infty }[({n}^{2}-n){a}_{n}+6n{a}_{n}+4{a}_{n}-(n+2)(n+1){a}_{n+2}]{x}^{n} & = & -4 \\ \sum _{n=0}^{\infty }[{n}^{2}{a}_{n}+5n{a}_{n}+4{a}_{n}-(n+2)(n+1){a}_{n+2}]{x}^{n} & = & -4 \\ \sum _{n=0}^{\infty }[({n}^{2}+5n+4){a}_{n}-(n+2)(n+1){a}_{n+2}]{x}^{n} & = & -4 \\ \sum _{n=0}^{\infty }[(n+4)(n+1){a}_{n}-(n+2)(n+1){a}_{n+2}]{x}^{n} & = & -4\end{array}\]
      Looking at the coefficients of each power of x, we see that the constant term must be equal to \(-4,\) and the coefficients of all other powers of x must be zero. Then, looking first at the constant term,
      \[\begin{array}{lll}4{a}_{0}-2{a}_{2} & = & -4 \\ {a}_{2} & = & 2{a}_{0}+2\ \text{(step 3).}\end{array}\]
      For \(n\ge 1,\) we have
      \[\begin{array}{lll}(n+4)(n+1){a}_{n}-(n+2)(n+1){a}_{n+2} & = & 0 \\ (n+1)[(n+4){a}_{n}-(n+2){a}_{n+2}] & = & 0.\end{array}\]
      Since \(n\ge 1,\) \(n+1\ne 0,\) we see that
      \[(n+4){a}_{n}-(n+2){a}_{n+2}=0\]
      and thus
      \[{a}_{n+2}=\frac{n+4}{n+2}\ {a}_{n}.\]
      For even values of n, we have
      \[\begin{array}{lll}{a}_{4} & = & \frac{6}{4}(2{a}_{0}+2)=3{a}_{0}+3 \\ {a}_{6} & = & \frac{8}{6}(3{a}_{0}+3)=4{a}_{0}+4 \\ & \vdots \text{.} & \end{array}\]
      In general, \({a}_{2k}=(k+1)({a}_{0}+1)\) (step 5).
      For odd values of n, we have
      \[\begin{array}{lll}{a}_{3} & = & \frac{5}{3}\ {a}_{1} \\ {a}_{5} & = & \frac{7}{5}\ {a}_{3}=\frac{7}{3}\ {a}_{1} \\ {a}_{7} & = & \frac{9}{7}\ {a}_{5}=\frac{9}{3}\ {a}_{1}=3{a}_{1} \\ & \vdots \text{.} & \end{array}\]
      In general, \({a}_{2k+1}=\frac{2k+3}{3}\ {a}_{1}\) (step 5 continued).
      Putting this together, we have
      \[y(x)=\sum _{k=0}^{\infty }(k+1)({a}_{0}+1){x}^{2k}+\sum _{k=0}^{\infty }(\frac{2k+3}{3}){a}_{1}{x}^{2k+1}\ \text{(step 6).}\]

2. Find a power series solution for the following differential equations.

   1. \({y}^{'}+2xy=0\)
   2. \((x+1){y}^{'}=3y\)
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   1. \(y(x)={a}_{0}\sum _{n=0}^{\infty }\frac{{(-1)}^{n}}{n!}{x}^{2n}={a}_{0}{e}^{\text{-}{x}^{2}}\)
   2. \(y(x)={a}_{0}{(x+1)}^{3}\)

3. Find a power series solution to the Bessel equation of order 0 and graph the solution.

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   The Bessel equation of order 0 is given by

   \[{x}^{2}{y}^{″}+x{y}^{'}+{x}^{2}y=0.\]

   We assume a solution of the form \(y=\sum _{n=0}^{\infty }{a}_{n}{x}^{n}.\) Then \({y}^{'}(x)=\sum _{n=1}^{\infty }n{a}_{n}{x}^{n-1}\) and \({y}^{″}(x)=\sum _{n=2}^{\infty }n(n-1){a}_{n}{x}^{n-2}.\) Substituting this into the differential equation, we get

   \[\begin{array}{lll} \\ \\ {x}^{2}\sum _{n=2}^{\infty }n(n-1){a}_{n}{x}^{n-2}+x\sum _{n=1}^{\infty }n{a}_{n}{x}^{n-1}+{x}^{2}\sum _{n=0}^{\infty }{a}_{n}{x}^{n}=0 & & \text{Substitution.} \\ \sum _{n=2}^{\infty }n(n-1){a}_{n}{x}^{n}+\sum _{n=1}^{\infty }n{a}_{n}{x}^{n}+\sum _{n=0}^{\infty }{a}_{n}{x}^{n+2}=0 & & \text{Bring external factors within sums.} \\ \sum _{n=2}^{\infty }n(n-1){a}_{n}{x}^{n}+\sum _{n=1}^{\infty }n{a}_{n}{x}^{n}+\sum _{n=2}^{\infty }{a}_{n-2}{x}^{n}=0 & & \text{Re-index third sum.} \\ \sum _{n=2}^{\infty }n(n-1){a}_{n}{x}^{n}+{a}_{1}x+\sum _{n=2}^{\infty }n{a}_{n}{x}^{n}+\sum _{n=2}^{\infty }{a}_{n-2}{x}^{n}=0 & & \text{Separate}\ n=1\ \text{term from second sum.} \\ {a}_{1}x+\sum _{n=2}^{\infty }[n(n-1){a}_{n}+n{a}_{n}+{a}_{n-2}]{x}^{n}=0 & & \text{Collect summation terms.} \\ {a}_{1}x+\sum _{n=2}^{\infty }[({n}^{2}-n){a}_{n}+n{a}_{n}+{a}_{n-2}]{x}^{n}=0 & & \text{Multiply through in first term.} \\ {a}_{1}x+\sum _{n=2}^{\infty }[{n}^{2}{a}_{n}+{a}_{n-2}]{x}^{n}=0. & & \text{Simplify.}\end{array}\]

   Then, \({a}_{1}=0,\) and for \(n\ge 2,\)

   \[\begin{array}{l}{n}^{2}{a}_{n}+{a}_{n-2}=0 \\ \\ {a}_{n}=-\frac{1}{{n}^{2}}\ {a}_{n-2}.\end{array}\]

   Because \({a}_{1}=0,\) all odd terms are zero. Then, for even values of n, we have

   \[\begin{array}{lll}{a}_{2} & = & -\frac{1}{{2}^{2}}\ {a}_{0} \\ {a}_{4} & = & -\frac{1}{{4}^{2}}\ {a}_{2}=\frac{1}{{4}^{2}⋅{2}^{2}}\ {a}_{0}. \\ {a}_{6} & = & -\frac{1}{{6}^{2}}\ {a}_{4}=-\frac{1}{{6}^{2}⋅{4}^{2}⋅{2}^{2}}\ {a}_{0}\end{array}\]

   In general,

   \[{a}_{2k}=\frac{{(-1)}^{k}}{{(2)}^{2k}{(k!)}^{2}}\ {a}_{0}.\]

   Thus, we have

   \[y(x)={a}_{0}\sum _{k=0}^{\infty }\frac{{(-1)}^{k}}{{(2)}^{2k}{(k!)}^{2}}{x}^{2k}.\]

   The graph appears below for \({a}_{0}=1\).

4. Verify that the expression found in is a solution to the Bessel equation of order 0.

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5. \({y}^{″}+6{y}^{'}=0\)

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6. \(5{y}^{″}+{y}^{'}=0\)

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   \(y={a}_{0}+5{a}_{1}\sum _{n=1}^{\infty }\frac{{(\text{-}x\text{/}5)}^{n}}{n!}={c}_{0}+5{c}_{1}{e}^{\text{-}x\text{/}5}\)

7. \({y}^{″}+25y=0\)

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8. \({y}^{″}-y=0\)

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   \(y={a}_{0}\sum _{n=0}^{\infty }\frac{{(x)}^{2n}}{(2n)!}+{a}_{1}\sum _{n=0}^{\infty }\frac{{(x)}^{2n+1}}{(2n+1)!}\)

9. \(2{y}^{'}+y=0\)

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10. \({y}^{'}-2xy=0\)

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    \(y={a}_{0}\sum _{n=0}^{\infty }\frac{{x}^{2n}}{n!}={c}_{0}{e}^{{x}^{2}}\)

11. \((x-7){y}^{'}+2y=0\)

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12. \({y}^{″}-x{y}^{'}-y=0\)

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    \(y={a}_{0}\sum _{n=0}^{\infty }\frac{{x}^{2n}}{{2}^{n}n!}+{a}_{1}\sum _{n=0}^{\infty }\frac{{x}^{2n+1}}{1⋅3⋅5⋅7\cdots (2n+1)}\)

13. \((1+{x}^{2}){y}^{″}-4x{y}^{'}+6y=0\)

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14. \({x}^{2}{y}^{″}-x{y}^{'}-3y=0\)

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    \(y={c}_{0}{x}^{3}+\frac{{c}_{1}}{x}\)

15. \({y}^{″}-8{y}^{'}=0,\ y(0)=-2,\ \ {y}^{'}(0)=10\)

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16. \({y}^{″}-2xy=0,\ y(0)=1,\ \ {y}^{'}(0)=-3\)

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    \(y=1-3x+\frac{2{x}^{3}}{3!}-\frac{12{x}^{4}}{4!}+\frac{16{x}^{6}}{6!}-\frac{120{x}^{7}}{7!}+\cdots\)

17. The differential equation \({x}^{2}{y}^{″}+x{y}^{'}+({x}^{2}-1)y=0\) is a Bessel equation of order 1. Use a power series of the form \(y=\sum _{n=0}^{\infty }{a}_{n}{x}^{n}\) to find the solution.

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18. If \(y\) and \(z\) are both solutions to \({y}^{″}+2{y}^{'}+y=0,\) then \(y+z\) is also a solution.

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    True

19. The following system of algebraic equations has a unique solution:

    \(\begin{array}{l}6{z}_{1}+3{z}_{2}=8 \\ 4{z}_{1}+2{z}_{2}=4.\end{array}\)

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20. \(y={e}^{x}\text{cos}(3x)+{e}^{x}\text{sin}(2x)\) is a solution to the second-order differential equation \({y}^{″}+2y'+10=0.\)

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    False

21. To find the particular solution to a second-order differential equation, you need one initial condition.

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22. \({y}^{″}-2y=0\)

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    second order, linear, homogeneous, \({\lambda }^{2}-2=0\)

23. \({y}^{″}-3y+2y=\text{cos}\ (t)\)

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24. \({(\frac{dy}{dt})}^{2}+y{y}^{'}=1\)

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    first order, nonlinear, nonhomogeneous

25. \(\frac{{d}^{2}y}{d{t}^{2}}+t\frac{dy}{dt}+{\text{sin}}^{2}(t)y={e}^{t}\)

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26. \({y}^{″}+9y=0\)

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    \(y={c}_{1}\text{cos}\ (3x)+{c}_{2}\text{sin}\ (3x)\)

27. \({y}^{″}+2{y}^{'}+y=0\)

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28. \({y}^{″}-2{y}^{'}+10y=4x\)

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    \(y={c}_{1}{e}^{x}\text{cos}\ (3x)+{c}_{2}{e}^{x}\text{sin}\ (3x)+\frac{2}{5}x+\frac{2}{25}\)

29. \({y}^{″}=\text{cos}\ (x)+2{y}^{'}+y\)

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30. \({y}^{''}+5y'+4y=x+{e}^{2x}\)

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    \(y={c}_{1}{e}^{\text{-}x}+{c}_{2}{e}^{-4x}+\frac{x}{4}+\frac{{e}^{2x}}{18}-\frac{5}{16}\)

31. \({y}^{″}=3{y}^{'}+x{e}^{\text{-}x}\)

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32. \({y}^{″}-{x}^{2}=-3{y}^{'}-\frac{9}{4}y+3x\)

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    \(y={c}_{1}{e}^{(-3\text{/}2)x}+{c}_{2}x{e}^{(-3\text{/}2)x}+\frac{4}{9}{x}^{2}+\frac{4}{27}x-\frac{16}{27}\)

33. \({y}^{″}=2\ \text{cos}\ x+{y}^{'}-y\)

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34. \({y}^{″}+4{y}^{'}+6y=0,\) \(y(0)=0,\) \({y}^{'}(0)=\sqrt{2}\)

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    \(y={e}^{-2x}\text{sin}\ (\sqrt{2}x)\)

35. \({y}^{″}=3y-\text{cos}\ (x),\) \(y(0)=\frac{9}{4},\) \({y}^{'}(0)=0\)

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36. \(4{y}^{'}=-6y+2{y}^{″},\) \(y(0)=0,\) \(y(1)=1\)

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    \(y=\frac{{e}^{1-x}}{{e}^{4}-1}({e}^{4x}-1)\)

37. \({y}^{″}=3x-y-{y}^{'},\) \(y(0)=-3,\) \(y(1)=0\)

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38. The motion of a swinging pendulum for small angles \(\theta\) can be approximated by \(\frac{{d}^{2}\theta }{d{t}^{2}}+\frac{g}{L}\theta =0,\) where \(\theta\) is the angle the pendulum makes with respect to a vertical line, g is the acceleration resulting from gravity, and L is the length of the pendulum. Find the equation describing the angle of the pendulum at time \(t,\) assuming an initial displacement of \({\theta }_{0}\) and an initial velocity of zero.

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    \(\theta (t)={\theta }_{0}\text{cos}\ (\sqrt{\frac{g}{l}}L)\)

39. Find the general solution to this equation (Hint: call \({\omega }_{0}=\sqrt{b\text{/}a}\)).

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40. Assuming the system starts from rest, show that the particular solution can be written as \(y=\frac{2}{a({\omega }_{0}{}^{2}-{\omega }^{2})}\ \text{sin}\ (\frac{{\omega }_{0}-\omega t}{2})\ \text{sin}\ (\frac{{\omega }_{0}+\omega t}{2}).\)

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