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 $2{\frac {d^{2}y}{dt^{2}}}\,{\frac {dy}{dt}}={\frac {d\left({\dfrac {dy}{dt}}\right)^{2}}{dt}},\quad \left({\frac {dy}{dt}}\right)^{2}+n^{2}(y^{2}-C^{2})=0$,

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

${\frac {dy}{dt}}=-n{\sqrt {y^{2}-C^{2}}}\quad {\text{and}}\quad {\frac {dy}{\sqrt {C^{2}-y^{2}}}}=n\cdot dt.$

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

${\frac {1}{\sqrt {C^{2}-y^{2}}}}={\frac {d(\arcsin {\dfrac {y}{C}})}{dy}};$

whence, passing from angles to sines,

$\arcsin {\frac {y}{C}}=nt+C_{1}\quad {\text{and}}\quad y=C\sin(nt+C_{1}),$