Practicing Success
If y2 = p(x), a polynomial of degree 3 then $2 \frac{d}{d x}\left(y^3 \frac{d^2 y}{d x^2}\right)$ is equal to : |
p'''(x) + p'(x) p'''(x) + p''(x) p(x) p'''(x) a constant |
p(x) p'''(x) |
$y^2=p(x) \Rightarrow 2 y y_1=p'(x)$ $\Rightarrow 2\left(y y_2+y_1 y_1\right)=p''(x)$ $\Rightarrow y y_2=\frac{1}{2}\left(p''(x)-2 y_1{ }^2\right)$ $\Rightarrow y^3 y_2=\frac{1}{2}\left(p''(x) y^2-2\left(y y_1\right)^2\right)$ $=\frac{1}{2} p''(x) . p(x)-\left(\frac{1}{2} p'(x)\right)^2$ $\Rightarrow \frac{d}{d x}\left(y^3 y_2\right)=\frac{1}{2}\left[p'''(x) p(x)+p'(x) p''(x)\right]-2 \times\left(\frac{1}{2} p'(x)\right) \times\left(\frac{1}{2} p''(x)\right)$ $\Rightarrow 2 \frac{d}{d x} \left(y^3 y_2\right)=p(x) p'''(x)$ Hence (3) is correct answer. |