Download App

CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY5 Marks
Question
If $\text{x}=\sin\text{t}$ and $\text{y}=\sin\text{pt},$ prove that $(1-\text{x}^2)\frac{\text{d}^2\text{y}}{\text{dx}^2}-\text{x}\frac{\text{dy}}{\text{dx}}+\text{p}^2\text{ y}=0.$
✓

Answer

We have, $\text{x}=\sin\text{t}$ and $\text{y}=\sin\text{pt}$
$\therefore\ \frac{\text{dx}}{\text{dt}}=\cos\text{t}$ and $\frac{\text{dy}}{\text{dt}}=\text{p}\cos\text{pt}$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}=\frac{\text{p}\cdot\cos\text{pt}}{\cos\text{t}}$
$\Rightarrow\ \text{y}'=\frac{\text{p}\cdot\cos\text{pt}}{\cos\text{t}}$
$\Rightarrow\ \text{y}'^2\cos^2\text{t}=\text{p}^2\cos^2\text{pt}$
$\Rightarrow\ \text{y}'^{2}(1-\sin^2\text{t})=\text{p}^2(1-\sin^2\text{pt})$
$\Rightarrow\ \text{y}'^2(1-\text{x}^2)=\text{p}^2(1-\text{y}^2)$
Differentiating above w.r.t. x, we get
$2\text{y}'\text{y}''(1-\text{x}^2)-2\text{xy}'^2=\text{p}^2(-2\text{yy}')$
$\Rightarrow\ \text{y}''(1-\text{x}^2)-\text{xy}'=-\text{p}^2\text{y}$
$\Rightarrow\ \text{y}''(1-\text{x}^2)-\text{xy}'+\text{p}^2\text{y}=0$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "5 Marks Questions" from CONTINUITY AND DIFFERENTIABILITY→CONTINUITY AND DIFFERENTIABILITY — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

Find the area of the region lying in the first quadrant and bounded by $y = 4x^2, x = 0, y = 1$ and $y = 4$.
→
A bag contains 10 balls, each marked with one of the digits from 0 to 9. If four balls are drawn successively with replacement from the bag, what is the probability that none is marked with the digit 0?
→
If $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ are non-zero, non-coplanar vectors, prove that the vector is coplanar:
$5\vec{\text{a}}+6\vec{\text{b}}+7\vec{\text{c}},\ 7\vec{\text{a}}-8\vec{\text{b}}+9\vec{\text{c}}$ and $3\vec{\text{a}}+20\vec{\text{b}}+5\vec{\text{c}}$
→
If the value of c prescribed in Roll's theorem for the function$\text{f}(\text{x})=2\text{x}(\text{x}-3)^{\text{n}}$ on the interval $\big[0,2\sqrt3\big]$ is $\frac{3}{4},$ write the value of n (a positive integers).
→
Prove that $\int\limits_{0}^{\text{a}}\text{f(x)dx}=\int\limits_{0}^{\text{a}}\text{f (a - x) dx.}$
Hence, evaluate $\int\limits_{0}^{\pi/2}\frac{\text{dx}}{\text{1 + tan x}}$.
→
If $\text{y}=\text{x}^3\log\text{x},$ Prove that $\frac{\text{d}^4\text{y}}{\text{dx}^4}=\frac{6}{\text{x}}$
→
Two bad eggs are accidently mixed up with ten good ones. Three eggs are drawn at random with replacement from this lot. Compute the mean for the number of bad eggs drawn.
→
Evaluate the following integrals:$\int\sin^{-1}\sqrt{\text{x}}\text{dx}$
→
Prove that: $\begin{vmatrix}(b+c)^2&a^2&{bc}\\(c+a)^2&{b}^2&{ca}\\(a+b)^2&{c}^2&{ab}\end{vmatrix}$ $=(a-b)(b-c)(c-b)(a+b+c)({a}^2+{b}^2+{c}^2)$
→
Find the distance of a point $(2,4,-1)$ from the line $\frac{x+5}{1}=\frac{y+3}{4}=\frac{z-6}{-9}$.
→