Q71 / 180 · KCET 2013 · Mathematics
If $m$ and $n$ are order and degree of the differential equation
$\left(y^{\prime \prime}\right)^{5}+4 \cdot \frac{\left(y^{\prime \prime}\right)^{3}}{y^{\prime \prime \prime}}+y^{\prime \prime \prime}=\sin x$, then
- A$m=3, n=5$
- B$m=3, n=1$
- C$m=3, n=3$
- Correct answerD$m=3, n=2$
Solution
Given differential equation is,
$\left(\frac{d^{2} y}{d x^{2}}\right)^{5}+4 \cdot \frac{\left(\frac{d^{2} y}{d x^{2}}\right)^{3}}{\left(\frac{d^{3} y}{d x^{3}}\right)}+\left(\frac{d^{3} y}{d x^{3}}\right)=\sin x$
$\Rightarrow\left(\frac{d^{2} y}{d x^{2}}\right)^{5} \cdot\left(\frac{d^{3} y}{d x^{3}}\right)+4 \cdot\left(\frac{d^{2} y}{d x^{2}}\right)^{3}+\left(\frac{d^{3} y}{d x^{3}}\right)^{2}$
$=\sin x \cdot\left(\frac{d^{3} y}{d x^{3}}\right)$
Now, $m=$ order $=$ highest order derivative $=3$
and $n=$ degree $=$ power of highest order derivative $=2$
Hence, the correct answer is option: 'D'.