The order and degree of the differential equation $\left(\frac{d^3 y}{d…
The order and degree of the differential equation $\left(\frac{d^3 y}{d x^3}\right)^{\frac{1}{2}}-2\left(\frac{d y}{d x}\right)^{\frac{1}{4}}+x y=0$ are respectively
3 and 12
3 and 2
3 and 4
3 and 6
Solution
Given, $y_3^{\frac{1}{2}}-2\left(y_1\right)^{\frac{1}{4}}+x y=0$
$
\Rightarrow\left(y_3^{\frac{1}{2}}-x y\right)^4=\left[2\left(y_1\right)^{\frac{1}{4}}\right]^4
$
$
\begin{aligned}
& \Rightarrow y_3^{\frac{1}{2}}\left[{ }^4 C_1\left(y_3\right) x y+{ }^4 C_3(x y)^3\right] \\
& =\left[16 y_1-{ }^4 C_0\left(y_3\right)^2-{ }^4 C_2\left(y_3\right)(x y)^2-{ }^4 C_4(x y)^4\right]
\end{aligned}
$
Squaring both side we have $=y_3\left[{ }^4 C_1\left(y_3\right) x y+{ }^4 C_3(x y)^3\right]^2=\left[16 y_1-{ }^4 C_0\left(y_3\right)^2-{ }^4 C_2\left(y_3\right)\right.$ $\left.(x y)^2-{ }^4 C_4(x y)^4\right]^2$ which is now free from fractions in the power of derivatives $\left(y_1, y_2, y_3\right)$.
Clearly, order $=3$
Since $y_3$ will be having power 4 in right side.
Hence degree $=4$