The degree of is not well defined.

\frac{d^{3} y}{{dx}^{3}} + \frac{dy}{dx} = e^{y}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

e^{(\frac{dy}{dx})} = k + \frac{dy}{dx}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\sqrt{(1 + {(\frac{dy}{dx})}^{2})} = y \frac{d^{3} y}{{dx}^{3}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{d^{3} y}{{dx}^{3}} + cosy \frac{dy}{dx} = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

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

Suggest Corrections

Similar questions

{(\frac{d^{2} y}{d x^{2}})}^{3} + {(\frac{d y}{d x})}^{2} + \sin (\frac{d y}{d x}) + 1 = 0

\frac{d^{3} y}{d x^{3}} + \frac{d^{2} y}{d x^{2}} + \frac{d y}{d x} + y \sin y = 0

\frac{d^{3} y}{d x^{3}} + \frac{d^{2} y}{d x^{2}} \cdot \frac{d y}{d x} = 0

\frac{d^{3} y}{d x^{3}} + {(\frac{d^{2} y}{d x^{2}})}^{3} + \frac{d y}{d x} + 4 y = \sin x

Join BYJU'S Learning Program