Solve the equation x²y′ = y - x - 1.

Question

Solve the equation [latex]x²y^{′} = y - x - 1.[/latex]

Accepted Answer

We make the usual substitutions [latex]y = \sum{c_{n} x^{n}} and y^{′} = \sum{n c_{n} x^{n-1}}[/latex], which yield

[latex]x² \sum\limits_{n=1}^{\infty}{n c_{n} x^{n-1}} = -1 - x + \sum\limits_{n=0}^{\infty}{c_{n}x^{n}}[/latex]

so that

[latex]\sum\limits_{n=1}^{\infty}{n c_{n} x^{n+1}} = -1 - x + \sum\limits_{n=0}^{\infty}{c_{n}x^{n}}[/latex].

Because of the presence of the two terms -1 and -x on the right-hand side, we need to split off the first two terms, [latex]c_{0} + c_{1} x[/latex], of the series on the right for comparison. If we also shift the index of summation on the left by -1 (replace n = 1 with n = 2 and n with n - 1), we get

[latex]\sum\limits_{n=2}^{\infty}{(n - 1)c_{n-1} x^{n}} = -1 - x + c_{0} + c_{1} x +\sum\limits_{n=2}^{\infty}{c_{n} x^{n}}.[/latex]

Because the left-hand side contains neither a constant term nor a term containing x to the first power, the identity principle now yields [latex]c_{0} = 1, c_{1} = 1, and c_{n} = (n - 1)c_{n-1 }[/latex] for n ≧ 2. It follows that

[latex]c_{2} = 1 · c_{1} = 1!, c_{3} = 2 · c_{2} = 2!, c_{4} = 3 · c_{3} = 3!,[/latex]

and, in general, that

[latex]c_{n} = (n - 1)![/latex] for n ≧ 2:

Thus we obtain the power series

[latex]y(x) = 1 + x + \sum\limits_{n=2}^{\infty}{(n - 1)! x^{n}}.[/latex]

But the radius of convergence of this series is

ρ [latex]= \underset{n → \infty}{\lim} \frac{(n - 1)!}{n!} = \underset{n → \infty}{\lim} \frac{1}{n} = 0.[/latex]

so the series converges only for x = 0. What does this mean? Simply that the given differential equation does not have a (convergent) power series solution of the assumed form [latex]y = \sum{c_{n}x^{n}}[/latex]. This example serves as a warning that the simple act of writing [latex]y = \sum{c_{n}x^{n}}[/latex] involves an assumption that may be false.

Question 11.1.3: Solve the equation x²y′ = y - x - 1.

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