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Prove properties of a function satisfying xy′ – y = x sin x

Define a function

    \[ s(x) = \frac{\sin x}{x} \]

for x \neq 0, and s(0) = 1. Define a second function

    \[ T(x) = \int_0^x s(t)\, dt. \]

Prove that the function f(x) = xT(x) satisfies the differential equation

    \[ xy' - y = x \sin x \qquad \text{on } (-\infty, +\infty).\]

Further, find all solutions to this differential equation on the interval. Prove that there is no solution such that the initial condition f(0) = 1 is satisfied. Why does this not contradict Theorem 8.3?


Proof. To prove that f(x) = xT(x) satisfies the given differential equation we use Theorem 8.3 to find a formula for all solutions to this differential equation. For x \neq 0 we have,

    \[ xy' - y = x \sin x \quad \implies \quad y' - \left( \frac{1}{x} \right) y = \sin x. \]

Therefore, using Theorem 8.3 we let

    \[ P(x) = -\frac{1}{x}, \qquad Q(x) = \sin x \]

giving us

    \[ A(x) = \int_1^x -\frac{1}{t} \, dt = -\log t \Bigr \rvert_1^x = -\log x. \]

Therefore,

    \begin{align*}  f(x) &= be^{\log x} + e^{\log x} \int_1^x (\sin t) e^{-\log t} \, dt \\  &= x \left( b + \int_0^x s(t) \, dt - \int_0^1 s(t) \, dt \right) \\  &= x \left(b + T(x) - T(1)\right) \\  &= xT(x) + Cx. \end{align*}

Hence, f(x) = xT(x) is a solution satisfying f(1) = C = b-T(1). Also, since f and x \sin x are both 0 when x = 0, f is valid on the entire interval (-\infty, +\infty).

Finally, the differential equation has no solution satisfying f(0) = 1 (since f(0) = 0 for all C). This does not contradict Theorem 8.3 since P(x) = \frac{1}{x} is not continuous on any interval around 0. \qquad \blacksquare

6 comments

    • RoRi says:

      Yeah, I think you’re right. We need b = T(1). I’ll try to fix it soon, but I want to think about it for a bit first in case I had some reason for thinking this was right that I don’t remember right now.

      • Another thing that I guess that you should do is to separate the differencial equation in two parts, one of then in the interval (0,\infty) and the other (-\infty,0), it because 1/x isn’t continuous between (\infty,\infty), after this u should show that the lateral limits of f'(x) were equal, showing the differenciability of f(x).

      • RoRi says:

        Yeah, I’ll need to look at it and fix the whole thing. My finals are starting in less than a week though, so I probably won’t be fixing anything until after those are over. Been committing so much time to the blog, that I’m behind on my actual classes! Thanks for pointing out the problems though, please keep commenting if you find mistakes in other problems (which, no doubt, you will).

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