The order and degree of the differential equa | vedclass

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Question

(A) Given the differential equation: $\rho = \frac{{\left[ {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} \right]^{3/2}}}{{\frac{{d^2y}}{{dx^2}}}}$Mul

Vedclass · Accepted Answer

$2, 2$

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(A) Given the differential equation: $\rho = \frac{{\left[ {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} \right]^{3/2}}}{{\frac{{d^2y}}{{dx^2}}}}$Multiply both sides by $\frac{{d^2y}}{{dx^2}}$:$\rho \cdot \frac{{d^2y}}{{dx^2}} = \left[ {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} \right]^{3/2}$To eliminate the fractional exponent,square both sides:$\left( \rho \cdot \frac{{d^2y}}{{dx^2}} \right)^2 = \left[ {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} \right]^3$The highest order derivative present is $\frac{{d^2y}}{{dx^2}}$,so the order is $2$.The power of the highest order derivative after rationalizing the equation is $2$,so the degree is $2$.Therefore,the order and degree are $2$ and $2$ respectively.

The order and degree of the differential equation $\rho = \frac{{\left[ {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} \right]^{3/2}}}{{\frac{{d^2y}}{{dx^2}}}}$ are respectively

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Reason $(R)$: An algebraic equation having two arbitrary constants is the general solution of a second-order differential equation.

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