<p><strong> If <math><mrow><mi>x</mi><mo>=</mo><mi>s</mi><mi>i</mi><mi>n</mi><mi>t</mi></mrow></math> and <math><mrow><mi>y</mi><mo>=</mo><mi>s</mi><mi>i</mi><mi>n</mi><mo> </mo><mi>p</mi><mi>t</mi></mrow></math> , prove that <math><mrow><mrow><mo>(</mo><mrow><mn>1</mn><mo>−</mo><msup><mrow><mi>x</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow><mo>)</mo></mrow><mfrac><mrow><msup><mrow><mi>d</mi></mrow><mrow><mn>2</mn></mrow></msup><mi>y</mi></mrow><mrow><mi>d</mi><msup><mrow><mi>x</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow></mfrac><mo>−</mo><mi>x</mi><mfrac><mrow><mi>d</mi><mi>y</mi></mrow><mrow><mi>d</mi><mi>x</mi></mrow></mfrac><mo>+</mo><mo> </mo><msup><mrow><mi>p</mi></mrow><mrow><mn>2</mn></mrow></msup><mi>y</mi><mo>=</mo><mn>0</mn></mrow></math></strong></p>

This is a Long Answer Type Questions as classified in NCERT ExemplarSol:Given that x=sint and y=sinptDifferentiating both sides w.r.t. t⇒ dxdt=cost and dydt=cospt.p=p.cospt⇒ dydx=dydtdxdt=p.cosptcost⇒ dydx=p.cosptcostAgain differentiating both sides w.r.t. x⇒ ddx(dydx)=p.ddx(cosptcost)⇒ d2ydx2=p.[cost.ddx(cospt)−cospt.ddx(cost)cos2t]⇒ d2ydx2=p.[cost.(−sinpt).pdtdx−cospt.(−sint).dtdxcos2t]⇒ d2ydx2=p.[−pcost.sinpt+cospt.sintcos2t]dtdx⇒ d2ydx2=p.[−pcost.sinpt+cospt.sintcos2t]1cost⇒ d2ydx2=p.(−pcost.sinpt+cospt.sintcos3t)Now we have to prove that (1−x2)d2ydx2−xdydx+ p2y=0L.H.S.=(1−x2)[p.(−pcost.sinpt+cospt.sintcos3t)]−x(p.cosptcost)+ p2y⇒ =(1−sin2t)[p.(−pcost.sinpt+cospt.sintcos3t)]−p.sintcosptcost+ p2.sinpt⇒ =cos2t[−p2cost.sinpt+pcospt.sintcos3t]−p.sintcosptcost+ p2.sinpt⇒ =−p2cost.sinpt+pcospt.sintcost−p.sintcosptcost+ p2.sinpt⇒ =−p2cost.sinpt+pcospt.sint−p.sintcospt+p2.sinptcostcost⇒ =0cost=0=R.H.S. Hence, proved.

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Continuity and Differentiability Ncert Solutions Maths class 12th Maths Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Five

If $x = s i n t$ and $y = s i n p t$ , prove that $(1 - x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} + p^{2} y = 0$

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alok kumar singh | Contributor-Level 10

3 months ago

This is a Long Answer Type Questions as classified in NCERT Exemplar

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$\begin{array}{l} G i v e n t h a t x = s i n t a n d y = s i n p t \\ D i f f e r e n t i a t i n g b o t h s i d e s w . r . t . t \\ \Rightarrow \frac{d x}{d t} = c o s t a n d \frac{d y}{d t} = c o s p t . p = p . c o s p t \\ \Rightarrow \frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{p . c o s p t}{c o s t} \\ \Rightarrow \frac{d y}{d x} = \frac{p . c o s p t}{c o s t} \\ A g a i n d i f f e r e n t i a t i n g b o t h s i d e s w . r . t . x \\ \Rightarrow \frac{d}{d x} (\frac{d y}{d x}) = p . \frac{d}{d x} (\frac{c o s p t}{c o s t}) \\ \Rightarrow \frac{d^{2} y}{d x^{2}} = p . [\frac{c o s t . \frac{d}{d x} (c o s p t) - c o s p t . \frac{d}{d x} (c o s t)}{c o s^{2} t}] \\ \Rightarrow \frac{d^{2} y}{d x^{2}} = p . [\frac{c o s t . (- s i n p t) . p \frac{d t}{d x} - c o s p t . (- s i n t) . \frac{d t}{d x}}{c o s^{2} t}] \\ \Rightarrow \frac{d^{2} y}{d x^{2}} = p . [\frac{- p c o s t . s i n p t + c o s p t . s i n t}{c o s^{2} t}] \frac{d t}{d x} \\ \Rightarrow \frac{d^{2} y}{d x^{2}} = p . [\frac{- p c o s t . s i n p t + c o s p t . s i n t}{c o s^{2} t}] \frac{1}{c o s t} \\ \Rightarrow \frac{d^{2} y}{d x^{2}} = p . (\frac{- p c o s t . s i n p t + c o s p t . s i n t}{c o s^{3} t}) \\ N o w w e h a v e t o p r o v e t h a t \\ (1 - x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} + p^{2} y = 0 \\ L . H . S . = (1 - x^{2}) [p . (\frac{- p c o s t . s i n p t + c o s p t . s i n t}{c o s^{3} t})] - x (\frac{p . c o s p t}{c o s t}) + p^{2} y \\ \Rightarrow = (1 - s i n^{2} t) [p . (\frac{- p c o s t . s i n p t + c o s p t . s i n t}{c o s^{3} t})] - \frac{p . s i n t c o s p t}{c o s t} + p^{2} . s i n p t \\ \Rightarrow = c o s^{2} t [\frac{- p^{2} c o s t . s i n p t + p c o s p t . s i n t}{c o s^{3} t}] - \frac{p . s i n t c o s p t}{c o s t} + p^{2} . s i n p t \\ \Rightarrow = \frac{- p^{2} c o s t . s i n p t + p c o s p t . s i n t}{c o s t} - \frac{p . s i n t c o s p t}{c o s t} + p^{2} . s i n p t \\ \Rightarrow = \frac{- p^{2} c o s t . s i n p t + p c o s p t . s i n t - p . s i n t c o s p t + p^{2} . s i n p t c o s t}{c o s t} \\ \Rightarrow = \frac{0}{c o s t} = 0 = R . H . S . \\ H e n c e, p r o v e d . \end{array}$

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If $x = s i n t$ and $y = s i n p t$ , prove that $(1 - x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} + p^{2} y = 0$

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If x=sint and y=sin pt , prove that (1−x2)d2ydx2−xdydx+ p2y=0

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If $x = s i n t$ and $y = s i n p t$ , prove that $(1 - x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} + p^{2} y = 0$