Search any question & find its solution
Question:
Answered & Verified by Expert
If $m$ and $n$ are order and degree of the equation $\left(\frac{d^{2} y}{d x^{2}}\right)^{5}+4 \cdot \frac{\left(\frac{d^{2} y}{d x^{2}}\right)^{3}}{\left(\frac{d^{3} y}{d x^{3}}\right)}+\left(\frac{d^{3} y}{d x^{3}}\right)=x^{2}-1$, then
Options:
Solution:
2167 Upvotes
Verified Answer
The correct answer is:
$m=3, n=2$
The given differential equation can be rewritten as
$$
\begin{array}{l}
\left(\frac{d^{2} y}{d x^{2}}\right)^{5} \cdot \frac{d^{3} y}{d x^{3}}+4\left(\frac{d^{2} y}{d x^{2}}\right)^{3}+\left(\frac{d^{3} y}{d x^{3}}\right)^{2} \\
=\left(x^{2}-1\right) \frac{d^{3} y}{d x^{3}} \Rightarrow \text { order }(m)=3
\end{array}
$$
and $\quad$ degree $(n)=2$
$$
\begin{array}{l}
\left(\frac{d^{2} y}{d x^{2}}\right)^{5} \cdot \frac{d^{3} y}{d x^{3}}+4\left(\frac{d^{2} y}{d x^{2}}\right)^{3}+\left(\frac{d^{3} y}{d x^{3}}\right)^{2} \\
=\left(x^{2}-1\right) \frac{d^{3} y}{d x^{3}} \Rightarrow \text { order }(m)=3
\end{array}
$$
and $\quad$ degree $(n)=2$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.