Question 8.3.4: Use infinite series to determine the solution to the initial...

Considering the series expansions for y, y′, and y″, we observe that

y = a_{0} +a_{1}t +a_{2}t^{2 }+a_{3}t^{3}+· · ·+a_{n}t^{n} +· · ·
y′ = a_{1} +2a_{2}t +3a_{3}t^{2} +4a_{4}t^{3}+· · ·+(n+1)a_{n+1}t^{n} +· · ·
y″= 2a_{2} +6a_{3}t +12a_{4}t^{2} +20a_{5}t^{3}+· · ·+(n+2)(n+1)a_{n+2}t^{n} +· · ·

From the differential equation y″ −2y′ −3y = 0, we know that y″ = 2y′ +3y. Equating like coefficients from the expressions for y and 2y′+3y, we find the recurrence relation

2a_{2} = 2a_{1} +3a_{0}
6a_{3} = 4a_{2} +3a_{1}
12a_{4} = 6a_{3} +3a_{2}
20a5 = 8a_{4} +3a_{3}

¦

More generally, we can state that for any n ≥ 2,

a_{n} = \frac{(2n−2)a_{n−1} +3a_{n−2}}{n(n−1)}

Using the given initial conditions, we find that a_{0} =y(0)=4  and  a_{1} =y′(0)=0, and subsequently that

a_{2} = \frac {2a_{1} +3a_{0}}{2}= \frac {0+12}{2}= 6
a_{3} = \frac {4a_{2} +3a_{1}}{6}= \frac {24+0}{6}= 4
a_{4} = \frac {6a_{3} +3a_{2}}{12}= \frac {24+18}{12}= \frac {7}{2}

and therefore the solution to the IVP is

y = 4+6t^{2} +4t^{3} + \frac {7}{2}t^{4}+· · ·

We can confirm that this is in fact the correct solution by solving the IVP through another approach and considering power series expansions of the basic solution functions. In particular, since the characteristic equation of (8.3.18) is r^{2} − 2r − 3 = 0 with roots r = 3 and r = −1, the general solution of the DE is

y = c_{1}e^{3t} +c_{2}e^{−t}

It is a standard exercise to show that the values of the constants that satisfy the initial conditions are c_{1} = 1 and c_{2} = 3, so that

y = e^{3t} +3e^{−t}

If we now employ the standard power series expansion for e^{t} to write series expansions for the two solutions present in y, and then combine like terms, we observe that

y = e^{3t} +3e^{−t}

=(1+3t + \frac {9t^{2}}{2!}+ \frac {27t^{3}}{3!}+ \frac {81t^{4}}{4!}+···)+(3−3t + \frac {3t^{2}}{2!}− \frac {3t^{3}}{3!}+ \frac {3t^{4}}{4!}−···)

=4+ \frac {12t^{2}}{2!}+ \frac {24t^{3}}{3!}+ \frac {84t^{4}}{4!}+···

=4+6t^{2}+4t^{3}+ \frac {7}{2}t^{4}+···

which is precisely the power series expansion of the solution we found at the outset.