Page:Calculus Made Easy.pdf/264

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

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

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

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 ${\displaystyle y=A\sin(pt+q)}$. However, let us go to work.

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

${\displaystyle C}$ being a constant. Then, taking the square roots,

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

whence, passing from angles to sines,