Question

If x = a sin pt, y = b cost p, then show that $(a^2-x^2)y\frac{d^{2}y}{d{x}^2}+b^2=0.$

Answer 1

Here x = a sin pt, y = b cos pt

Differentiating w.r.t. t,

$\frac{dx}{dt}=apcospt,\frac{dy}{dt}=-bpsinpt\Rightarrow \frac{dy}{dx}=-\frac{b}{a}tanpt$

On differentiating w.r.t. x, we get :$\frac{d^{2}y}{d{x}^2}=-\frac{bpse{c}^{2}pt}{a}\times \frac{dt}{dx}$

$\Rightarrow \frac{d^{2}y}{d{x}^2}=-\frac{bpse{c}^{2}pt}{a}\times \frac{1}{apcospt}=-\frac{b}{a^2}\times \frac{1}{co{s}^{2}pt}\times \frac{1}{cospt}$

$\Rightarrow \frac{d^{2}y}{d{x}^2}=-\frac{b}{a^2}\times \frac{1}{1-si{n}^{2}pt}\times \frac{b}{y}\Rightarrow \frac{d^{2}y}{d{x}^2}=-\frac{b}{a^2}\times \frac{1}{1-\frac{x^2}{a^2}}\times \frac{b}{y}$

$\Rightarrow \frac{d^{2}y}{d{x}^2}=-\frac{b}{a^2-x^2}\times \frac{b}{y}\Rightarrow (a^2-x^2)y\frac{d^{2}y}{d{x}^2}=-b^2$

$\therefore (a^2-x^2)y\frac{d^{2}y}{d{x}^2}+b^2=0.$