If y = (cos^{-1} x)^2,then (1-x^2) y_2 + p x | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given $y = (\cos^{-1} x)^2$. Differentiating with respect to $x$,we get $y_1 = 2(\cos^{-1} x) \cdot \left(-\frac{1}{\sqrt{1-x^2}}\right)$. So,$\sq

Vedclass · Accepted Answer

-$3$

Vedclass · Answer

(C) Given $y = (\cos^{-1} x)^2$. Differentiating with respect to $x$,we get $y_1 = 2(\cos^{-1} x) \cdot \left(-\frac{1}{\sqrt{1-x^2}}ight)$. So,$\sqrt{1-x^2} y_1 = -2 \cos^{-1} x$. Squaring both sides,we get $(1-x^2) y_1^2 = 4 (\cos^{-1} x)^2 = 4y$. Differentiating again with respect to $x$,we get $(1-x^2) \cdot 2y_1 y_2 + y_1^2 (-2x) = 4y_1$. Dividing by $2y_1$ (assuming $y_1 
eq 0$),we get $(1-x^2) y_2 - x y_1 = 2$. Thus,$(1-x^2) y_2 - x y_1 - 2 = 0$. Comparing this with $(1-x^2) y_2 + p x y_1 + q = 0$,we get $p = -1$ and $q = -2$. Therefore,$p+q = -1 + (-2) = -3$.

If $y = (\cos^{-1} x)^2$,then $(1-x^2) y_2 + p x y_1 + q = 0$. Find the value of $p+q$.

Similar Questions

Let $p(x)$ be a polynomial such that $p(x) - p'(x) = x^n$,where $n$ is a positive integer. Then,$p(0)$ equals

If $y = (x^2 - 1)^m$,then the $(2m)^{th}$ differential coefficient of $y$ is

If the function $y = \sin^{-1} x$,then $\left(1-x^2\right) \frac{d^2 y}{d x^2}$ is equal to

If $y=e^{m \sin ^{-1} x}$, then $(1-x^{2}) \frac{d^{2} y}{d x^{2}}-x \frac{d y}{d x}-k y=0$, where $k$ is equal to

If $y=ae^x+be^{-x}+c$, where $a, b, c$ are parameters, then $y^{\prime \prime \prime}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module