If order and degree of the differential equation $\left(\frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}\right)^5+4…
If order and degree of the differential equation $\left(\frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}\right)^5+4 \frac{\left(\frac{\mathrm{~d}^2 y}{\mathrm{~d} x^2}\right)^5}{\left(\frac{\mathrm{~d}^3 y}{\mathrm{~d} x^3}\right)}+\frac{\mathrm{d}^3 y}{\mathrm{~d} x^3}=\sin x$, are m and n respectively, then the value of $\left(m^2+n^2\right)$ is equal to
29
13
5
8
Solution
Given differential equation is
$\left(\frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}\right)^5+4 \frac{\left(\frac{\mathrm{~d}^2 y}{\mathrm{~d} x^2}\right)^5}{\frac{\mathrm{~d}^3 y}{\mathrm{~d} x^3}}+\frac{\mathrm{d}^3 y}{\mathrm{~d} x^3}=\sin x$
$\left(\frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}\right)^5 \cdot \frac{\mathrm{~d}^3 y}{\mathrm{~d} x^3}+4\left(\frac{\mathrm{~d}^2 y}{\mathrm{~d} x^2}\right)^5+\left(\frac{\mathrm{d}^3 y}{\mathrm{~d} x^3}\right)^2=\sin x \frac{\mathrm{~d}^3 y}{\mathrm{~d} x^3}$
Here, order $=3$ and degree $=2$
$\begin{aligned}
& \therefore \quad m=3, n=2 \\
& \therefore \quad m^2+n^2=3^2+2^2=13
\end{aligned}$