Page:Calculus Made Easy.pdf/264

 Example 5. Let $$\quad \dfrac{d^2 y}{dt^2} + n^2 y = 0$$.

In this case we have a differential equation of the second degree, in which $$y$$ appears in the form of a second differential coefficient, as well as in person.

Transposing, we have $$\dfrac{d^2 y}{dt^2} = - n^2 y$$.

It appears from this that we have to do with a function such that its second differential coefficient is proportional to itself, but with reversed sign. In Chapter XV. we found that there was such a function–namely, the sine (or the cosine also) which possessed this property. So, without further ado, we may infer that the solution will be of the form $$y = A \sin (pt + q)$$. However, let us go to work.

Multiply both sides of the original equation by $$2\dfrac{dy}{dt}$$ and integrate, giving us $$2\dfrac{d^2 y}{dt^2}\, \dfrac{dy}{dt} + 2x^2 y \dfrac{dy}{dt} = 0$$, and, as $$2 \frac{d^2y}{dt^2}\, \frac{dy}{dt} = \frac{d \left(\dfrac{dy}{dt}\right)^2}{dt},\quad \left(\frac{dy}{dt}\right)^2 + n^2 (y^2-C^2) = 0$$,

$$C$$ being a constant. Then, taking the square roots,

But it can be shown that (see p. 171)

whence, passing from angles to sines,