The order and degree of the differential equa | vedclass

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Question

(B) Given differential equation is $\sqrt{\frac{dy}{dx}}-4 \frac{dy}{dx}-7x=0$. Rearranging the terms,we get $\sqrt{\frac{dy}{dx}}=4 \frac{dy}{dx}+7x$

Vedclass · Accepted Answer

$1$ and $2$

Vedclass · Answer

(B) Given differential equation is $\sqrt{\frac{dy}{dx}}-4 \frac{dy}{dx}-7x=0$. Rearranging the terms,we get $\sqrt{\frac{dy}{dx}}=4 \frac{dy}{dx}+7x$. Squaring both sides to eliminate the radical,we obtain $\frac{dy}{dx} = (4 \frac{dy}{dx} + 7x)^2$. Expanding the right side,we get $\frac{dy}{dx} = 16(\frac{dy}{dx})^2 + 56x(\frac{dy}{dx}) + 49x^2$. The highest order derivative present is $\frac{dy}{dx}$,so the order is $1$. The highest power of the highest order derivative is $2$,so the degree is $2$. Thus,the order is $1$ and the degree is $2$.

The order and degree of the differential equation $\sqrt{\frac{dy}{dx}}-4 \frac{dy}{dx}-7x=0$ are respectively.

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