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If the order and degree of the differential equation d2ydx2=(1+dydx)13\sqrt{\frac{d^2y}{dx^2}} = \left(1 + \frac{dy}{dx}\right)^{\frac{1}{3}} are aa and bb respectively, then the value of a2+b2a^2 + b^2 is

Solution

✅ Correct Option: 4

To clear fractional powers, raise both sides to the power 6: (d2ydx2)3=(1+dydx)2\left(\frac{d^2y}{dx^2}\right)^3 = \left(1 + \frac{dy}{dx}\right)^2. Highest order derivative is d2ydx2\frac{d^2y}{dx^2} with power 3. So order a=2a = 2 and degree b=3b = 3. Therefore a2+b2=4+9=13a^2 + b^2 = 4 + 9 = 13.

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