KCET · Maths · Application of Derivatives
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
- C 5
- D 2
Answer & Solution
Correct Answer
(C) 5
Step-by-step Solution
Detailed explanation
Given differential equation is
\[
\begin{gathered}
\left(1+\left(\frac{d y}{d x}\right)^{2}\right)^{2 / 3}=\left(\frac{d^{2} y}{d x^{2}}\right) \\
\Rightarrow \quad\left(1+\left(\frac{d y}{d x}\right)^{2}\right)^{2}=\left(\frac{d^{2} y}{d x^{2}}\right)^{3} \\
\text { Now, } \quad \text { Order }=n=2 \\
\text { Degree }=m=3 \\
\text { Then, } \quad \frac{m+n}{m-n}=\frac{3+2}{3-2}=5
\end{gathered}
\]
\[
\begin{gathered}
\left(1+\left(\frac{d y}{d x}\right)^{2}\right)^{2 / 3}=\left(\frac{d^{2} y}{d x^{2}}\right) \\
\Rightarrow \quad\left(1+\left(\frac{d y}{d x}\right)^{2}\right)^{2}=\left(\frac{d^{2} y}{d x^{2}}\right)^{3} \\
\text { Now, } \quad \text { Order }=n=2 \\
\text { Degree }=m=3 \\
\text { Then, } \quad \frac{m+n}{m-n}=\frac{3+2}{3-2}=5
\end{gathered}
\]
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