Question 11.2.6: Solve the initial value problem (t² - 2t - 3)d²y/dt² + 3(t -...
Solve the initial value problem
(t² – 2t – 3)\frac{d²y}{dt²} + 3(t – 1)\frac{dy}{dt} + y = 0; y(1) = 4, y^{′}(1) = -1. (14)
We need a general solution of the form \sum{c_{n}(t – 1)^{n}}. But instead of substituting this series in (14) to determine the coefficients, it simplifies the computations if we first make the substitution x = t – 1, so that we wind up looking for a series of the form \sum{c_{n} x^{n}} after all. To transform Eq. (14) into one with the new independent variable x, we note that
t² – 2t – 3 = (x + 1)² – 2(x + 1) – 3 = x² – 4,
\frac{dy}{dt} = \frac{dy}{dx} \frac{dx}{dt} = \frac{dy}{dx} = y^{′},
and
\frac{d²y}{dt²} = [\frac{d}{dx}(\frac{dy}{dx})]\frac{dx}{dt} = \frac{d}{dx} (y^{′}) = y^{″},
where primes denote differentiation with respect to x. Hence we transform Eq. (14) into
(x² – 4)y^{″} + 3xy^{′} + y = 0
with initial conditions y = 4 and y^{′} = 1 at x = 0 (corresponding to t = 1). This is the initial value problem we solved in Example 5, so the particular solution in (12) is available. We substitute t – 1 for x in Eq. (12) and thereby obtain the desired particular solution
y(x) = 4 + x + \frac{1}{2} x^{2} + \frac{1}{6}x^{3} + \frac{3}{32}x^{4} + \frac{1}{30}x^{5}+ · · · · . (12)
y(t) = 4 + (t – 1) + \frac{1}{2} (t – 1)^{2} + \frac{1}{6}(t – 1)^{3} + \frac{3}{32}(t – 1)^{4} + \frac{1}{30}(t – 1)^{5}+ · · · · .
This series converges if -1 < t < 3. (Why?) A series such as this can be used to estimate numerical values of the solution. For instance,
y(0.8) = 4 + (-0.2) + \frac{1}{2} (-0.2)^{2} + \frac{1}{6}(-0.2)^{3} + \frac{3}{32}(-0.2)^{4} + \frac{1}{30}(-0.2)^{5} + · · · ,so that y(0.8) ≈ 3.8188.