Miscellaneous · Q8: $ye^{x/y}\,dx = (xe^{x/y} + y^2)\,dy$

Basics you need

Solution

1. Write the equation as $\frac{dx}{dy} = F(x, y)$

   Divide both sides by $ye^{x/y}\,dy$⁠, which is not 0 because $y \ne 0$ and $e^{x/y} > 0$⁠:

   $$\begin{aligned} \frac{dx}{dy} &= \frac{xe^{x/y} + y^2}{ye^{x/y}} \\ &= \frac{x}{y} + ye^{-x/y} = F(x, y).\quad\ldots(1) \end{aligned}$$
   We choose $\frac{dx}{dy}$ because the divisor $ye^{x/y}$ is never 0, while the $\frac{dy}{dx}$ form would need $xe^{x/y} + y^2 \ne 0$⁠. Splitting the fraction: $\frac{xe^{x/y}}{ye^{x/y}} = \frac{x}{y}$ and $\frac{y^2}{ye^{x/y}} = ye^{-x/y}$⁠.
   Section 9.4.2, Note: the form $\frac{dx}{dy} = F(x, y)$⁠. Example 12 uses the same form

2. Test for homogeneity: it fails

   For $\lambda \ne 0$⁠, $\frac{\lambda x}{\lambda y} = \frac{x}{y}$⁠, so

   $$F(\lambda x, \lambda y) = \frac{x}{y} + \lambda\,ye^{-x/y}.$$

   This is not $F(x, y)$ unless $\lambda = 1$ (for example at $x = 0$⁠, $y = 1$⁠: $F(0, \lambda) = \lambda$⁠, but $F(0, 1) = 1$⁠). So the equation is not homogeneous.

   The terms $ye^{x/y}\,dx$ and $xe^{x/y}\,dy$ have degree 1 outside the exponential, but $y^2\,dy$ has degree 2, and that mismatch breaks homogeneity. We still try $x = vy$⁠, because everything else depends only on $\frac{x}{y}$⁠.
   Section 9.4.2: definition of a homogeneous function. Basics: scale x and y together

3. Substitute $x = vy$

   Put $x = vy$⁠, so $\frac{x}{y} = v$ and, by the product rule (differentiating with respect to $y$⁠), $\frac{dx}{dy} = v + y\frac{dv}{dy}$⁠. Then (1) becomes

   $$v + y\frac{dv}{dy} = v + ye^{-v}.$$
   Here $y$ is the independent variable and $x$⁠, hence $v$⁠, depends on it. The first term $\frac{x}{y}$ of $F$ becomes $v$⁠, the second $ye^{-v}$⁠.
   Section 9.4.2, Note: $x = vy$⁠. Lesson: homogeneous equations

4. Simplify to $\frac{dv}{dy} = e^{-v}$

   The $v$ on each side cancels: $y\frac{dv}{dy} = ye^{-v}$⁠. Dividing by $y \ne 0$⁠,

   $$\frac{dv}{dy} = e^{-v}.$$
   For a homogeneous equation the right side would become a function of $v$ alone. Here an extra factor $y$ appears, but it is the same factor that multiplies $\frac{dv}{dy}$⁠, so it cancels. The result is even simpler than usual: $y$ has gone completely.
   Simplifying as in Section 9.4.2, equation (4). Lesson: Miscellaneous exercise player

5. Separate and integrate, with one constant

   Multiply by $e^v\,dy$⁠:

   $$\begin{aligned} e^v\,dv &= dy, \\ \int e^v\,dv &= \int dy, \\ e^v &= y + C, \end{aligned}$$

   where $C$ is an arbitrary constant.

   Now $v$ is on the left only and $y$ on the right only. Since $e^v > 0$⁠, we never divide by 0. One constant suffices for a first order equation.
   Section 9.4.1; Section 7.2, formula (xi). Basics: every method ends with an integral

6. Put $v = \frac{x}{y}$ back: the general solution

   $$e^{x/y} = y + C.$$

   It lives where $y \ne 0$ and $y + C > 0$ (an exponential is positive); there it gives $x = y\log(y + C)$⁠.

   The answer must be in the original variables. Taking logarithms gives the explicit form $\frac{x}{y} = \log(y + C)$⁠, which shows the domain restriction plainly.
   $e^u > 0$ and $\log$ is its inverse (Chapter 5, Section 5.4). Basics: e^{kt} and log

7. Check by substituting back

   Differentiate $e^{x/y} = y + C$ with respect to $y$⁠, with $x$ a function of $y$ (quotient rule for $\frac{x}{y}$⁠):

   $$\begin{aligned} e^{x/y}\cdot\frac{y\frac{dx}{dy} - x}{y^2} &= 1, \\ ye^{x/y}\frac{dx}{dy} - xe^{x/y} &= y^2, \\ ye^{x/y}\,dx &= \left(xe^{x/y} + y^2\right)dy, \end{aligned}$$

   which is the given equation, for every $C$⁠.

   Getting the equation back proves the answer. The same lines read backwards give a second solution: $e^{x/y}\cdot\frac{y\,dx - x\,dy}{y^2} = dy$ says $d\left(e^{x/y}\right) = dy$⁠, which integrates at once.
   Implicit differentiation, quotient rule, $(e^u)' = e^u u'$ (Chapter 5). Basics: a rule for slopes

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