Page:Elements of the Differential and Integral Calculus - Granville - Revised.djvu/175

Therefore

(B)	${\frac {d^{2}y}{dx^{2}}}={\frac {{\frac {dx}{dt}}{\frac {d^{2}y}{dt^{2}}}-{\frac {dy}{dt}}{\frac {d^{2}x}{dt^{2}}}}{\left({\frac {dx}{dt}}\right)^{3}}}$;

and so on for higher derivatives. This transformation is called changing the independent variable from x to t. It is usually better to work out examples by the methods illustrated above rather than by using the formulas deduced.

Illustrative Example 1. Change the independent variable from x to t in the equation.

(C)	$x^{2}{\frac {d^{2}y}{dx^{2}}}+x{\frac {dy}{dx}}+y=0$
(D)	$\ x=e^{t}\$.

Solution.

	${\frac {dx}{dt}}=e^{t}\$, therefore
(E)	${\frac {dt}{dx}}=e^{-t}\$.
Also	${\frac {dy}{dx}}={\frac {dy}{dt}}{\frac {dt}{dx}}$; therefore
(F)	${\frac {dy}{dx}}=e^{-t}{\frac {dy}{dt}}$
Also	${\frac {d^{2}y}{dx^{2}}}=e^{-t}{\frac {d}{dx}}\left({\frac {dy}{dt}}\right)-{\frac {dy}{dt}}e^{-t}{\frac {dt}{dx}}=e^{-t}{\frac {d}{dt}}\left({\frac {dy}{dt}}\right){\frac {dt}{dx}}-{\frac {dy}{dt}}e^{-t}{\frac {dt}{dx}}.$

Substituting in the last result from (E),

(G)	${\frac {d^{2}y}{dx^{2}}}=e^{-2t}{\frac {d^{2}y}{dt^{2}}}-{\frac {dy}{dt}}e^{-2t};$

Substituting (D), (F), (G) in (C),

$e^{2t}\left(e^{-2t}{\frac {d^{2}y}{dt^{2}}}-{\frac {dy}{dt}}e^{-2t}\right)+e^{t}\left(e^{-t}{\frac {dy}{dt}}\right)+y=0;$

and reducing, we get

${\frac {d^{2}y}{dt^{2}}}+y=0\$ Ans.

Since the formulas deduced in the Differential Calculus generally involve derivatives of y with respect to x, such formulas as (A) and (B) are especially useful when the parametric equations of a curve are given. Such examples were given in §66, and many others will be employed in what follows.