Consider the following differential equations | vedclass

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(C) For $D_1: y=4 \frac{dy}{dx}+3x \frac{dx}{dy}$. Multiplying by $\frac{dy}{dx}$, we get $y \frac{dy}{dx}=4(\frac{dy}{dx})^2+3x$. The highest derivat

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$2:3$

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(C) For $D_1: y=4 \frac{dy}{dx}+3x \frac{dx}{dy}$. Multiplying by $\frac{dy}{dx}$, we get $y \frac{dy}{dx}=4(\frac{dy}{dx})^2+3x$. The highest derivative is $\frac{dy}{dx}$, so order is $1$. The power of the highest derivative is $2$, so degree is $2$.For $D_2: \frac{d^2y}{dx^2}=(3+(\frac{dy}{dx})^2)^{\frac{4}{3}}$. Cubing both sides, we get $(\frac{d^2y}{dx^2})^3=(3+(\frac{dy}{dx})^2)^4$. The highest derivative is $\frac{d^2y}{dx^2}$, so order is $2$. The power of the highest derivative is $3$, so degree is $3$.For $D_3: [1+(\frac{dy}{dx})]^2=(\frac{dy}{dx})^2$. Expanding, $1+(\frac{dy}{dx})^2+2\frac{dy}{dx}=(\frac{dy}{dx})^2$, which simplifies to $1+2\frac{dy}{dx}=0$. The highest derivative is $\frac{dy}{dx}$, so order is $1$. The power of the highest derivative is $1$, so degree is $1$.Sum of orders $= 1+2+1 = 4$.Sum of degrees $= 2+3+1 = 6$.Required ratio $= \frac{4}{6} = \frac{2}{3}$.

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