⚠ ye^{xy}dx+xe^{xy}dy = 0 is read as the differential equation y' = (-ye^{xy})/(xe^{xy}): divide by dx and isolate y', the derivative of y with respect to x■ye is read as the product y*e, where e is the exponential constant: for the variable ye, write ye_■xy is read as the product x*y: for the variable xy, write xy_■xe is read as the product x*e, where e is the exponential constant: for the variable xe, write xe_■solve(y'=(-y*e^{x*y})/(x*e^{x*y}),x) is read as solve(y'=(-y*e^(x*y))/(x*e^(x*y)),x): use ^(...) to group an exponent, use * to multiply

1) Solve the separable equation $\mathrm{y\text{'}}=\frac{-y·e^{xy}}{x{e}^{xy}}$, where $y$ is a function of $x$

⇥1.1) Let's evaluate the domain of $-\frac{1}{x}$

⇥ For the expression $-\frac{1}{x}$ to be defined, we must have: $x\ne 0$

⇥Solution obtained: $x\ne 0$

The function of $x$ is defined for $x\ne 0$

Before dividing, $y=0$ gives the equilibrium solution $y=0$

On intervals where $y\ne 0$, separate the variables: $\frac{\mathrm{dy}}{y}=-\frac{\mathrm{dx}}{x}$

⇥1.2) Let's compute the integral: $\int \frac{1}{y}dy$

⇥ⓘ The derivative of $\ln \left|y\right|$ is $\frac{1}{y}$, so $\ln \left|y\right|$ is an antiderivative of $\frac{1}{y}$: $\int \frac{1}{y}dy$ = $\ln \left|y\right|$

⇥ $\ln \left|y\right|$

⇥1.3) Let's compute the integral: $\int (-\frac{1}{x})dx$

⇥ⓘ The derivative of $-\ln \left|x\right|$ is $-\frac{1}{x}$, so $-\ln \left|x\right|$ is an antiderivative of $-\frac{1}{x}$: $\int (-\frac{1}{x})dx$ = $-\ln \left|x\right|$

⇥ $-\ln \left|x\right|$

Integrating both sides gives $\ln \left|y\right|=-\ln \left|x\right|+C$, where $C$ is an arbitrary real constant

Taking the exponential gives $\left|y\right|=\exp \left(C\right)\exp (-\ln \left|x\right|)$ = $\frac{\exp \left(C\right)}{\left|x\right|}$, hence $y=\frac{K}{\left|x\right|}$ with $K$ = ±$\exp \left(C\right)$, a nonzero real constant

On each interval where $x\ne 0$, $\frac{1}{\left|x\right|}$ is $\frac{1}{x}$ or its opposite: this sign goes into $K$, $y=\frac{K}{x}$

$K=0$ gives back the equilibrium solution $y=0$: the solutions are $y=\frac{K}{x}$, where $K$ is an arbitrary real constant

▶Solution:   $y=\frac{K}{x}$