Question

The degree of is not well defined.

• A. $\frac{d^{3}y}{{\mathrm{dx}}^3}+\frac{\mathrm{dy}}{\mathrm{dx}}=e^y$
• B. $e^{\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}=k+\frac{\mathrm{dy}}{\mathrm{dx}}$
• C. $\sqrt{\left(1+{\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^2\right)}=y\frac{d^{3}y}{{\mathrm{dx}}^3}$
• D. $\frac{d^{3}y}{{\mathrm{dx}}^3}+\mathrm{cosy}\frac{\mathrm{dy}}{\mathrm{dx}}=0$

Answer 1

The correct option is B $e^{\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}=k+\frac{\mathrm{dy}}{\mathrm{dx}}$

The degree is clearly defined for the cases $\frac{d^{3}y}{{\mathrm{dx}}^3}+\mathrm{cosy}\frac{\mathrm{dy}}{\mathrm{dx}}=0$ and $\frac{d^{3}y}{{\mathrm{dx}}^3}+\frac{\mathrm{dy}}{\mathrm{dx}}=e^y$.
The DE $\sqrt{\left(1+{\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^2\right)}=y\frac{d^{3}y}{{\mathrm{dx}}^3}$ can be squared on both sides and the degree becomes 2.
Whereas in $e^{\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}=k+\frac{\mathrm{dy}}{\mathrm{dx}}$ even after applying ln on both sides we can't reduce to polynomial form.