Differential Equations Question 238
Question: If $ y^{2}=p(x) $ is a polynomial of degree 3, then what is $ 2\frac{d}{dx}[ y^{3}\frac{d^{2}y}{dx^{2}} ] $ equal to-
Options:
A) p’(x)p"’(x)
B) p"(x)p’"(x)
C) p(x)p"’(x)
D) A constant
Show Answer
Answer:
Correct Answer: C
Solution:
[c] Given that $ y^{2}=p(x) $ Differentiating
$ \Rightarrow 2yy_1=p’(x) $
$ [ herey_1=\frac{dy}{dx} ] $
$ \Rightarrow 2y_1=\frac{p’(x)}{y} $ Differentiating again,
$ \Rightarrow 2y_2=\frac{yp’’(x)-p’(x)y_1}{y^{2}},[ y_2=\frac{d^{2}y}{dx^{2}} ] $
$ \Rightarrow 2y_2=\frac{yp’’(x)-\frac{p’(x).p’(x)}{2y}}{y^{2}} $ $ =\frac{2y^{2}p’’(x)-p’(x){{)}^{2}}}{2y^{3}} $
$ \Rightarrow 2y^{3}y_2=\frac{1}{2}[2y^{2}p’’(x)-{{(p’(x))}^{2}}] $
$ \Rightarrow 2y^{3}y_2=\frac{1}{2}[2p(x)p’’(x)-{{(p’(x))}^{2}}] $
$ \Rightarrow 2\frac{d}{dx}(y^{3}y_2) $ $ =\frac{1}{2}[2p’(x)p’’(x)+2p(x)p’’’(x)-2p’(x)p’’(x)] $ $ =p(x)p’’’(x) $
BETA
