Example 20: $\log\frac{dy}{dx} = 3x + 4y$⁠, $y = 0$ when $x = 0$

Basics you need

Solution

1. Remove the log

   $$\frac{dy}{dx} = e^{3x + 4y}.$$
   $\log a = b$ means $a = e^b$⁠. The log in the question needs $\frac{dy}{dx} > 0$⁠, and $e^{3x + 4y}$ is always positive, so nothing is lost or gained.
   $\log$ and $e$ are inverse functions (Chapter 5, Section 5.4). Basics: exponentials

2. Split the exponential

   $$\frac{dy}{dx} = e^{3x}\cdot e^{4y}.\quad\ldots(1)$$
   $e^{p + q} = e^p\cdot e^q$⁠. Now the right side is a function of $x$ times a function of $y$⁠: the equation is separable.
   Section 9.4.1: $\frac{dy}{dx} = h(y)\cdot g(x)$⁠. Lesson: separate, then integrate

3. Separate the variables

   $$\begin{aligned} \frac{dy}{e^{4y}} &= e^{3x}\,dx, \quad\text{that is} \\ e^{-4y}\,dy &= e^{3x}\,dx. \end{aligned}$$
   Divide by $e^{4y}$⁠, which is never 0, so no solution is lost. $\frac{1}{e^{4y}} = e^{-4y}$⁠.
   Separating, Section 9.4.1. Basics: dy = y′ dx

4. Integrate, with one constant

   $$\begin{aligned} \int e^{-4y}\,dy &= \int e^{3x}\,dx \\ \frac{e^{-4y}}{-4} &= \frac{e^{3x}}{3} + C. \end{aligned}$$
   $\int e^{ky}\,dy = \frac{e^{ky}}{k}$⁠, here with $k = -4$ and $k = 3$⁠. The two constants of integration combine into one arbitrary $C$⁠.
   Section 7.3.1 and Section 7.2, formula (xi). Basics: every method ends with an integral

5. Clear the fractions

   Multiply by $-12$⁠:

   $$\begin{aligned} 3e^{-4y} &= -4e^{3x} - 12C \\ 4e^{3x} + 3e^{-4y} + 12C &= 0.\quad\ldots(2) \end{aligned}$$
   $12$ is the least common multiple of $4$ and $3$⁠, and the minus sign makes the $e^{-4y}$ term positive. This is the book's form of the general solution; $12C$ is still one arbitrary constant.
   The general solution (Section 9.3). Basics: one constant for order 1

6. Use the condition: $y = 0$ when $x = 0$

   $$\begin{aligned} 4e^0 + 3e^0 + 12C &= 0 \\ \Rightarrow\ 4 + 3 + 12C &= 0 \\ \Rightarrow\ C &= -\frac{7}{12}. \end{aligned}$$

   Putting $12C = -7$ in (2):

   $$4e^{3x} + 3e^{-4y} - 7 = 0.$$
   $e^0 = 1$⁠. The point $(0, 0)$ picks out the one member of the family (2) with $C = -\frac{7}{12}$⁠.
   Section 9.3: a particular solution. Lesson: general and particular solutions

7. Where the solution lives

   Solving for $y$⁠: $e^{-4y} = \frac{7 - 4e^{3x}}{3}$⁠, which must be positive. So $4e^{3x} < 7$⁠, that is $x < \frac13\log\frac74 \approx 0.18654$⁠, and there

   $$y = -\frac14\log\left(\frac{7 - 4e^{3x}}{3}\right).$$
   An exponential is always positive, so the right side must be too. As $x$ rises to $\frac13\log\frac74$⁠, the bracket falls to 0 and $y \to +\infty$⁠; as $x \to -\infty$⁠, $e^{3x} \to 0$ and $y \to -\frac14\log\frac73 \approx -0.21182$⁠.
   $\log$ is increasing and defined for positive numbers (Chapter 5, Section 5.4). Basics: exponentials

8. Check by substituting back, and see the curve

   Differentiate $4e^{3x} + 3e^{-4y} - 7 = 0$ with respect to $x$⁠:

   $$\begin{aligned} 12e^{3x} - 12e^{-4y}\,\frac{dy}{dx} &= 0 \\ \frac{dy}{dx} = \frac{e^{3x}}{e^{-4y}} &= e^{3x + 4y}, \end{aligned}$$

   so $\log\frac{dy}{dx} = 3x + 4y$⁠. At $(0, 0)$⁠: $4 + 3 - 7 = 0$⁠.

   Getting the equation back proves the answer. Every slope $e^{3x + 4y}$ is positive, so every solution rises; this one rises so fast that it reaches $+\infty$ at a finite $x$⁠.
   Implicit differentiation and the chain rule (Chapter 5, Sections 5.3.1 and 5.3.2). Basics: a rule for slopes

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