Q114 / 180 · KCET 2011 · Mathematics

If $m$ and $n$ are degree and order of $\left(1+y_{1}^{2}\right)^{2 / 3}=y_{2}$, then the value of $\frac{m+n}{m-n}$ is

  • A
    3
  • B
    4
  • Correct answer
    C
    5
  • D
    2

Solution

Given differential equation is $$ \begin{gathered} \left(1+\left(\frac{d y}{d x}\right)^{2}\right)^{2 / 3}=\left(\frac{d^{2} y}{d x^{2}}\right) \\ \Rightarrow \quad\left(1+\left(\frac{d y}{d x}\right)^{2}\right)^{2}=\left(\frac{d^{2} y}{d x^{2}}\right)^{3} \\ \text { Now, } \quad \text { Order }=n=2 \\ \text { Degree }=m=3 \\ \text { Then, } \quad \frac{m+n}{m-n}=\frac{3+2}{3-2}=5 \end{gathered} $$

Hence, the correct answer is option: 'C'.

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