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The order and degree of the differential equation $\frac{d^2 y}{d x^2}+y+\left(\frac{d y}{d x}-\frac{d^3 y}{d x^3}\right)^{3 / 2}=0$, are respectively.
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Verified Answer
The correct answer is:
3, 3
Given differential equation,
$\begin{aligned}
\frac{d^2 y}{d x^2}+y+\left(\frac{d y}{d x}-\frac{d^3 y}{d x^3}\right)^{3 / 2} & =0 \\
\left(\frac{d y}{d x}-\frac{d^3 y}{d x^3}\right)^{3 / 2} & =-\left(\frac{d^2 y}{d x^2}+y\right)
\end{aligned}$
$\left(\frac{d y}{d x}-\frac{d^3 y}{d x^3}\right)^3=\left(\frac{d^2 y}{d x^2}+y\right)^2$
Clearly, order and degree is 3 and 3 respectively.
$\begin{aligned}
\frac{d^2 y}{d x^2}+y+\left(\frac{d y}{d x}-\frac{d^3 y}{d x^3}\right)^{3 / 2} & =0 \\
\left(\frac{d y}{d x}-\frac{d^3 y}{d x^3}\right)^{3 / 2} & =-\left(\frac{d^2 y}{d x^2}+y\right)
\end{aligned}$
$\left(\frac{d y}{d x}-\frac{d^3 y}{d x^3}\right)^3=\left(\frac{d^2 y}{d x^2}+y\right)^2$
Clearly, order and degree is 3 and 3 respectively.
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