# How shall I proceed in this sum? If x=sint and y=sin2t, prove that (i) (1-x^2)(dy/dx)^2=4(1-y^2) (ii) (1-x^2)(d^2(y)/dx^2)-xdy/dx+4y=0

Mar 18, 2018

Given that $x = \sin t$ and $y = \sin 2 t$

#### Explanation:

Use $\frac{\mathrm{dy}}{\mathrm{dx}} = \left(\frac{\mathrm{dy}}{\mathrm{dt}}\right) . \left(\frac{\mathrm{dt}}{\mathrm{dx}}\right)$..........Chain Rule of differentiation

Here $\frac{\mathrm{dy}}{\mathrm{dt}} = 2 \cos 2 t$ and $\frac{\mathrm{dx}}{\mathrm{dt}} = \cos t$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 \cos 2 t}{\cos} t$

NOTE :
$\rightarrow \frac{\mathrm{dt}}{\mathrm{dx}} = \frac{1}{\frac{\mathrm{dx}}{\mathrm{dt}}}$

$\rightarrow 1 - {\left(\sin t\right)}^{2} = {\left(\cos t\right)}^{2}$

$\rightarrow \sin 2 t = 2 \sin t \cos t$

$\rightarrow \cos 2 t = 2 {\left(\cos t\right)}^{2} - 1$

For part (i) :-

Take LHS :-

$\Rightarrow \left(1 - {x}^{2}\right) {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2} = \left(1 - {\left(\sin t\right)}^{2}\right) . {\left[\frac{2 \cos 2 t}{\cos} t\right]}^{2}$

$\Rightarrow \left(1 - {x}^{2}\right) {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2} = \cancel{{\left(\cos t\right)}^{2}} . \frac{4 {\left(\cos 2 t\right)}^{2}}{\cancel{{\left(\cos t\right)}^{2}}} = 4 {\left(\cos 2 t\right)}^{2}$

Now take RHS :-

$4 \left(1 - {y}^{2}\right) = 4 \left(1 - {\left(\sin 2 t\right)}^{2}\right) = 4 {\left(\cos 2 t\right)}^{2}$

Thus,

LHS $=$ RHS $= 4 {\left(\cos 2 t\right)}^{2}$

For part (ii) I am calculating only $\frac{{d}^{2} \left(y\right)}{\mathrm{dx}} ^ 2$ so you plug the values in the equation and similarly proceed as done in part (i)

{because yahaan type karna bahut lamba pad raha hai ðŸ˜…ðŸ˜…ðŸ˜…}:-

$\frac{{d}^{2} \left(y\right)}{\mathrm{dx}} ^ 2 = \frac{d \left[\frac{\mathrm{dy}}{\mathrm{dx}}\right]}{\mathrm{dx}}$

$\Rightarrow \frac{{d}^{2} \left(y\right)}{\mathrm{dx}} ^ 2 = \frac{d \left[\frac{2 \cos 2 t}{\cos} t\right]}{\mathrm{dt}}$

rArr(d^2(y))/dx^2= [cost(-4sin2t)-2cos2t.(-sint)]/(cost)^2

$\Rightarrow \frac{{d}^{2} \left(y\right)}{\mathrm{dx}} ^ 2 = \frac{- 4 \cos t \left(2 \sin t \cos t\right) + 2 \left(2 {\left(\cos t\right)}^{2} - 1\right) \sin t}{\cos t} ^ 2$

$\Rightarrow \frac{{d}^{2} \left(y\right)}{\mathrm{dx}} ^ 2 = \frac{- 8 \sin t . {\left(\cos t\right)}^{2} + 4 \sin t . {\left(\cos t\right)}^{2} - 2 \sin t}{\cos t} ^ 2$

$\Rightarrow \frac{{d}^{2} \left(y\right)}{\mathrm{dx}} ^ 2 = \frac{- 4 \sin t . {\left(\cos t\right)}^{2} - 2 \sin t}{\cos t} ^ 2$

$\therefore \frac{{d}^{2} \left(y\right)}{\mathrm{dx}} ^ 2 = - 4 \sin t - 2 \sec t . \tan t$

Mar 18, 2018

$x = \sin \left(t\right)$

$\mathrm{dx} = \cos \left(t\right) \mathrm{dt}$

$y = \sin \left(2 t\right)$

$\mathrm{dy} = 2 \cos \left(2 t\right) \mathrm{dt}$

${d}^{2} y = - 4 \sin \left(2 t\right) {\mathrm{dt}}^{2}$

(i)

$\left(1 - {x}^{2}\right) {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2} = \left(1 - {\sin}^{2} \left(t\right)\right) {\left(\frac{2 \cos \left(2 t\right) \mathrm{dt}}{\cos \left(t\right) \mathrm{dt}}\right)}^{2}$

$= \cancel{{\cos}^{2} \left(t\right)} \cdot \frac{4 {\cos}^{2} \left(2 t\right) \cancel{\mathrm{dt}}}{\cancel{{\cos}^{2} \left(t\right)}} \cancel{\mathrm{dt}}$

$= 4 \left({\cos}^{2} \left(2 t\right)\right) = 4 \left(1 - {\sin}^{2} \left(2 t\right)\right) = 4 \left(1 - {y}^{2}\right)$

(ii)

$\left(1 - {x}^{2}\right) \left(\frac{{d}^{2} y}{\mathrm{dx}} ^ 2\right) - x \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) + 4 y$

$= \left(1 - {\sin}^{2} \left(t\right)\right) \left(\frac{- 4 \sin \left(2 t\right) {\mathrm{dt}}^{2}}{\cos \left(t\right) \mathrm{dt}} ^ 2\right) - \sin \left(t\right) \frac{2 \cos \left(2 t\right) \mathrm{dt}}{\cos \left(t\right) \mathrm{dt}} + 4 \sin \left(2 t\right)$

$= {\cos}^{2} \left(t\right) \left(\frac{- 4 \sin \left(2 t\right) {\mathrm{dt}}^{2}}{{\cos}^{2} \left(t\right) {\mathrm{dt}}^{2}}\right) - \sin \left(t\right) \frac{2 \cos \left(2 t\right) \mathrm{dt}}{\cos \left(t\right) \mathrm{dt}} + 4 \sin \left(2 t\right)$

$= \cancel{{\cos}^{2} \left(t\right)} \left(\frac{- 4 \sin \left(2 t\right) \cancel{{\mathrm{dt}}^{2}}}{\cancel{{\cos}^{2} \left(t\right)} \cancel{{\mathrm{dt}}^{2}}}\right) - \sin \left(t\right) \frac{2 \cos \left(2 t\right) \cancel{\mathrm{dt}}}{\cos \left(t\right) \cancel{\mathrm{dt}}} + 4 \sin \left(2 t\right)$

=cancel(-4sin(2t)) + cancel(4sin(2t)) - ???

Still in progress...