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If $x=\sin t$ and $y=\sin pt$, prove that $(1-x^2)\large \frac{d^2y}{dx^2}$$-x\large\frac{dy}{dx}+\normalsize p^2y=0$

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  • To find $\large\frac{dy}{dx}$ in the case of parametric functions,if $x=\phi(t)$ and $y=\psi(t)$,then $\large\frac{dy}{dx}$$=\large\frac{\Large\frac{dy}{dt}}{\Large\frac{dx}{dt}}$
Step 1:
$x=\sin t$ and $y=\sin pt$
Differentiating the above equation w.r.t $x$ we get,
$\large\frac{dx}{dt}$$=\cos t$
$\large\frac{dy}{dt}$$=p\cos pt$
Therefore $\large\frac{dy}{dx}$$=\large\frac{p\cos pt}{\cos t}$
Step 2:
Again differentiating w.r.t $x$ we get,
$\Rightarrow \large\frac{d}{dt}\big(\large\frac{p\cos pt}{\cos t}\big)\times \large\frac{1}{\cos t}$
$\Rightarrow \large\frac{\cos t(-p^2\sin pt)-p\cos pt.(-\sin t)}{\cos^2t}\times \large\frac{1}{\cos t}$
On simplifying we get,
$\large\frac{d^2y}{dx^2}=\large\frac{-p^2\cos t.\sin pt+p\cos pt.\sin t}{\cos^3t}$
Step 3:
$(1-x^2)\large\frac{d^2y}{dx^2}=$$(1-\sin^2t)\bigg[\large\frac{p\cos pt.\sin t-p^2\cos t\sin pt}{\cos^3t}\bigg]$
But $1-\sin^2t=\cos^2t$
$\Rightarrow \cos^2t\bigg[\large\frac{p\cos pt.\sin t-p^2\cos t.\sin pt}{\cos^3 t}\bigg]$
$\Rightarrow \large\frac{p\cos pt.\sin t-p^2\cos t.\sin pt}{\cos t}$------(1)
$x.\large\frac{dy}{dx}=\large\frac{\sin t.p\cos pt}{\cos t}$-----(2)
$p^2y=p^2\sin pt$------(3)
Step 4:
Now combing equ(1),(2) and (3) in
$\large\frac{p\cos pt.\sin t-p^2\cos t.\sin pt}{\cos t}-\frac{\sin t.p\cos pt}{\cos t}+$$p^2\sin pt$
On simplifying we get,
$\large\frac{p\cos pt.\sin t-p^2\cos t.\sin pt-\sin t.p\cos pt+p^2\sin pt.\cos t}{\cos t}$
$\Rightarrow 0$
Hence proved.
answered Jul 4, 2013 by sreemathi.v
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