$\frac{d y}{d x} + e^{y} = 0$

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Solution

$\frac{d y}{d x} + e^{y} = 0$

In this differential equation, the order of the highest order derivative is 1 and its power is 1. So, the order of the differential equation is 1 and its degree is 1.

It is a non-linear differential equation, as the exponent of the dependent variable is not equal to 1 (as per expansion series of $e^{y}$ ).

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Similar questions

If e^{x} + e^{y} = e^{x + y}, prove that \frac{d y}{d x} = - \frac{e^{x} (e^{y} - 1)}{e^{y} (e^{x} - 1)} or \frac{d y}{d x} + e^{y - x} = 0

e^{x} + e^{y} = e^{x + y}

\frac{d y}{d x} + e^{y - x} = 0

e^{x} + e^{y} = e^{x + y}

\frac{d y}{d x} = - \frac{e^{x} (e^{y} - 1)}{e^{y} (e^{x} - 1)}

e^{y} = y^{x},

\frac{d y}{d x} = \frac{{(\log y)}^{2}}{\log y - 1}

\frac{d y}{d x} + \frac{\cos x \sin y}{\cos y} = 0

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