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Find the order and degree of the following differential equation: $\large\frac{d^{2}y}{dx^{2}}-y+(\large\frac{dy}{dx}+\frac{d^{3}y}{dx^{3}})^{\frac{3}{2}}$=$0$

$\begin{array}{1 1}order =2; degree =1 \\ order =1; degree =2\\ order =2; degree =2\\ order =3; degree =3 \end{array}$

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• The order of a D.E is the order of the highest order derivative occurring in it. The degree of the D.E is the degree of the highest order derivative which occurs in it, after the D.E has been made free from radicals and fractions as far as derivatives are concerned.
Given $\large\frac{d^2y}{dx^2}$$-y+\bigg(\large\frac{dy}{dx}+\frac{d^3y}{dx^3}\bigg)^{3/2}=0 \large\frac{d^2y}{dx^2}$$-y=-\bigg(\large\frac{dy}{dx}+\frac{d^3y}{dx^3}\bigg)^{3/2}$
Squaring both sides
$\bigg(\large\frac{d^2y}{dx^2}\bigg)^2-$$2y \large\frac{d^2 y}{dx^2}$$+y^2=\bigg(\large\frac{dy}{dx}+ \frac{d^3 y}{dx^3}\bigg)^3$
$\bigg(\large\frac{d^3y}{dx^3}+\frac{dy}{dx}\bigg)^3-\bigg(\frac{d^2y}{dx^2}\bigg)$$+2y \large\frac{d^2 y}{dx^2}$$-y^2=0$
Order = 3
Degree = 3

answered Sep 3, 2013 by