The differential equation left[frac{1+left(fr | vedclass

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Question

(A) Given the differential equation: $\left[\frac{1+\left(\frac{dy}{dx}\right)^2}{\left(\frac{d^2y}{dx^2}\right)^{\frac{3}{2}}}\right]^2 = kx$ First,s

Vedclass · Accepted Answer

order $= 2$,degree $= 3$

Vedclass · Answer

(A) Given the differential equation: $\left[\frac{1+\left(\frac{dy}{dx}\right)^2}{\left(\frac{d^2y}{dx^2}\right)^{\frac{3}{2}}}\right]^2 = kx$ First,simplify the expression by removing the fractional exponent: $\frac{\left[1+\left(\frac{dy}{dx}\right)^2\right]^2}{\left(\frac{d^2y}{dx^2}\right)^3} = kx$ Now,multiply both sides by $\left(\frac{d^2y}{dx^2}\right)^3$: $\left[1+\left(\frac{dy}{dx}\right)^2\right]^2 = kx \left(\frac{d^2y}{dx^2}\right)^3$ The order of a differential equation is the highest derivative present,which is $\frac{d^2y}{dx^2}$,so the order is $2$. The degree is the power of the highest derivative when the equation is expressed as a polynomial in derivatives,which is $3$. Therefore,the order is $2$ and the degree is $3$.

The differential equation $\left[\frac{1+\left(\frac{dy}{dx}\right)^2}{\left(\frac{d^2y}{dx^2}\right)^{\frac{3}{2}}}\right]^2 = kx$ is of

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