The order and degree of the differential equa | vedclass

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(D) Given differential equation is $\sqrt{\frac{dy}{dx}} - 4 \frac{dy}{dx} - 7x = 0$. To find the degree,we must eliminate the radical sign. Rewrite t

Vedclass · Accepted Answer

$1$ and $2$

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(D) Given differential equation is $\sqrt{\frac{dy}{dx}} - 4 \frac{dy}{dx} - 7x = 0$. To find the degree,we must eliminate the radical sign. Rewrite the equation as $\sqrt{\frac{dy}{dx}} = 4 \frac{dy}{dx} + 7x$. Squaring both sides,we get: $(\frac{dy}{dx}) = (4 \frac{dy}{dx} + 7x)^2$. $(\frac{dy}{dx}) = 16(\frac{dy}{dx})^2 + 49x^2 + 56x \frac{dy}{dx}$. Rearranging the terms,we get $16(\frac{dy}{dx})^2 + (56x - 1)\frac{dy}{dx} + 49x^2 = 0$. 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

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