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The sum of order and degree of the differential equation {1+(dydx)2}52d2ydx2=p\frac{\left\{1+\left(\frac{dy}{dx}\right)^2\right\}^{\frac{5}{2}}}{\frac{d^2y}{dx^2}} = p is :

Solution

✅ Correct Option: 3

Rewrite as {1+(y′)2}5/2=p⋅y′′\left\{1+(y')^2\right\}^{5/2} = p \cdot y''.

Squaring: {1+(y′)2}5=p2(y′′)2\{1+(y')^2\}^5 = p^2 (y'')^2.

Order =2= 2, Degree =2= 2.

Sum =4= 4.

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