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  • If  then <img alt="\left(x^2-\frac{1}{27}\right) \frac{d^2 y}{d x^2}+x \frac{d y}{d x}="

If y^3-y=2 x then \left(x^2-\frac{1}{27}\right) \frac{d^2 y}{d x^2}+x \frac{d y}{d x}=

Option: 1

y


Option: 2

\frac{y}{3}


Option: 3

\frac{y}{9}


Option: 4

\frac{y}{27}


Answers (1)

best_answer

We have, 

y^3-y=2 x

Differentiating both sides w.r.t x, we get

\left(3 y^2-1\right) \frac{d y}{d x}=2 \Rightarrow \frac{d y}{d x}=\frac{2}{3 y^2-1}

Again, differentiating both sides w.r.t. x, we get

\frac{d^2 y}{d x^2}=\frac{-2 \cdot 6 y \frac{d y}{d x}}{\left(3 y^2-1\right)^2}

Using (1), we get

\frac{d^2 y}{d x^2}=\frac{-24 y}{\left(3 y^2-1\right)^3}

Now, 

\begin{aligned} & \left(x^2-\frac{1}{27}\right) \frac{d^2 y}{d x^2}+\frac{x d y}{d x} \\ \\& =\left(x^2-\frac{1}{27}\right)\left(\frac{-24 y}{\left(3 y^2-1\right)^3}\right)+\frac{2 x}{\left(3 y^2-1\right)} \\ \\& =\left(\frac{y^2\left(y^2-1\right)^2}{4}-\frac{1}{27}\right)\left(\frac{-24 y}{\left(3 y^2-1\right)^3}\right)+\frac{y\left(y^2-1\right)}{3 y^2-1} \\ \end{aligned}

                                                                                         \left(\because y^3-y=2 x\right)

\begin{aligned} & =\frac{\left\{27 y^2\left(y^2-1\right)^2-4\right\}}{108} \frac{(-24 y)}{\left(3 y^2-1\right)^3}+\frac{y\left(y^2-1\right)}{3 y^2-1} \\ & =\frac{y}{9}\left\{\frac{-54 y^2\left(y^2-1\right)^2+8}{\left(3 y^2-1\right)^3}+\frac{9\left(y^2-1\right)}{3 y^2-1}\right\} \\ & =\frac{y}{9}\left\{\frac{-2(1+\alpha)(\alpha-2)^2+8}{\alpha^3}\right\}+\frac{3(\alpha-2)}{\alpha} \\ \end{aligned}

                                                                          \text { where } \alpha=3 y^2-1

=\frac{y}{9}

Posted by

shivangi.bhatnagar

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