If y=eacos-1x,-1≤x≤1,show that (1-x2)\(\frac{d^2y}{dx^2}\)-x\(\frac{dy}{dx}\)-a2y=0
If y=eacos-1x,-1≤x≤1,show that (1-x2)\(\frac{d^2y}{dx^2}\)-x\(\frac{dy}{dx}\)-a2y=0
Solution and Explanation
It is given that,y=eacos-1x
Taking logarithms on both sides, we obtain
logy=acos-1xloge
logy=acos-1x
Differentiating both sides with respect to x, we obtain
\(\frac{1}{y}\)\(\frac{dy}{dx}\)=a.\(-\frac{1}{\sqrt{1-x^2}}\)
\(\Rightarrow\) \(\frac{dy}{dx}\)=-a\(\frac{y^2}{{1-x^2}}\)
By squaring both sides, we obtain
(\(\frac{dy}{dx}\))2=a2\(\frac{y^2}{1-x^2}\)
Again differentiating both sides with respect to x,we obtain
(1-x2)\(\frac{d^2y}{dx^2}\)-x\(\frac{dy}{dx}\)-a2y=0
Hence,proved
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Concepts Used:
Continuity & Differentiability
Definition of Differentiability
f(x) is said to be differentiable at the point x = a, if the derivative f ‘(a) be at every point in its domain. It is given by
Definition of Continuity
Mathematically, a function is said to be continuous at a point x = a, if
It is implicit that if the left-hand limit (L.H.L), right-hand limit (R.H.L), and the value of the function at x=a exist and these parameters are equal to each other, then the function f is said to be continuous at x=a.
If the function is unspecified or does not exist, then we say that the function is discontinuous.