Page:Calculus Made Easy.pdf/265

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FINDING SOLUTIONS

245

where $C_{1}$ is a constant angle that comes in by integration.

Or, preferably, this may be written $y=A\sin nt+B\cos nt$ , which is the solution.

Example 6. ${\dfrac {d^{2}y}{dt^{2}}}-n^{2}y=0$ .

Here we have obviously to deal with a function $y$ which is such that its second differential coefficient is proportional to itself. The only function we know that has this property is the exponential function (see p. 143), and we may be certain therefore that the solution of the equation will be of that form.

Proceeding as before, by multiplying through by $2{\dfrac {dy}{dx}}$ , and integrating, we get $2{\dfrac {d^{2}y}{dx^{2}}}\,{\dfrac {dy}{dx}}-2x^{2}y{\dfrac {dy}{dx}}=0$ , and, as

${\begin{aligned}2{\frac {d^{2}y}{dx^{2}}}\,{\frac {dy}{dx}}={\frac {d\left({\dfrac {dy}{dx}}\right)^{2}}{dx}},\quad \left({\frac {dy}{dx}}\right)^{2}-n^{2}(y^{2}+c^{2})=0,\\{\frac {dy}{dx}}-n{\sqrt {y^{2}+c^{2}}}=0,\end{aligned}}$

where $c$ is a constant, and ${\dfrac {dy}{\sqrt {y^{2}+c^{2}}}}=n\,dx$ . Now, if

$\quad w=\log _{\epsilon }(y+{\sqrt {y^{2}+c^{2}}})=\log _{\epsilon }u$ ,
${\frac {dw}{du}}={\frac {1}{u}},\quad {\frac {du}{dy}}=1+{\frac {y}{\sqrt {y^{2}+c^{2}}}}={\frac {y+{\sqrt {y^{2}+c^{2}}}}{\sqrt {y^{2}+c^{2}}}}$

and

${\frac {dw}{dy}}={\frac {1}{\sqrt {y^{2}+c^{2}}}}$ .

Hence, integrating, this gives us

$\log _{\epsilon }(y+{\sqrt {y^{2}+c^{2}}})=nx+\log _{\epsilon }C,$
$y+{\sqrt {y^{2}+c^{2}}}=C\epsilon ^{nx}.....................{\text{(1)}}$

Now

$\qquad (y+{\sqrt {y^{2}+c^{2}}})\times (-y+{\sqrt {y^{2}+c^{2}}})=c^{2}$ ;

whence

$\qquad -y+{\sqrt {y^{2}+c^{2}}}={\dfrac {c^{2}}{C}}\epsilon ^{-nx}..............{\text{(2)}}$