Miscellaneous · Q10: $\left[\frac{e^{-2\sqrt x}}{\sqrt x} - \frac{y}{\sqrt x}\right]\frac{dx}{dy} = 1$

Basics you need

Solution

1. Turn the equation over

   The bracket times $\frac{dx}{dy}$ equals 1, so $\frac{dx}{dy} \ne 0$⁠, and

   $$\frac{dy}{dx} = \frac{1}{dx/dy} = \frac{e^{-2\sqrt x}}{\sqrt x} - \frac{y}{\sqrt x}.$$
   A product equal to 1 has no zero factor, so $\frac{dx}{dy} \ne 0$ and the bracket equals its reciprocal $\frac{dy}{dx}$⁠. Because of $\sqrt x$ in a denominator, the equation needs $x > 0$⁠; the book's "$(x \ne 0)$" together with $\sqrt x$ means exactly this. The equation has order 1 and, as a polynomial in $\frac{dy}{dx}$⁠, degree 1. (Strictly, the given form also needs $\frac{dy}{dx} \ne 0$⁠; the $\frac{dy}{dx}$ form below holds at every $x > 0$⁠.)
   Reciprocal derivatives; Sections 9.2.1 and 9.2.2. Basics: what a derivative means

2. Recognise the linear form: $P = \frac{1}{\sqrt x}$⁠, $Q = \frac{e^{-2\sqrt x}}{\sqrt x}$

   $$\frac{dy}{dx} + \frac{1}{\sqrt x}\,y = \frac{e^{-2\sqrt x}}{\sqrt x}.$$
   Move $-\frac{y}{\sqrt x}$ to the left. Now $y$ and $\frac{dy}{dx}$ appear only to the first power and are not multiplied together, and $P = \frac{1}{\sqrt x}$⁠, $Q = \frac{e^{-2\sqrt x}}{\sqrt x}$ depend on $x$ only: the form $\frac{dy}{dx} + Py = Q$ of Section 9.4.3. Both are defined for every $x > 0$⁠.
   Section 9.4.3, step (i). Lesson: linear equations

3. The integrating factor: $e^{2\sqrt x}$

   $$\text{I.F.} = e^{\int\frac{1}{\sqrt x}\,dx} = e^{\int x^{-\frac{1}{2}}\,dx} = e^{2\sqrt x}.$$
   By the power rule with $n = -\frac{1}{2}$⁠: $\int x^{-\frac{1}{2}}\,dx = \frac{x^{\frac{1}{2}}}{\frac{1}{2}} = 2\sqrt x$⁠. Any antiderivative will do for the I.F., so no constant is added here.
   Section 9.4.3, step (ii); Section 7.2, formula (i). Basics: exponentials

4. Multiply by $e^{2\sqrt x}$⁠: the left side is one derivative

   $$\begin{aligned} e^{2\sqrt x}\frac{dy}{dx} + \frac{e^{2\sqrt x}}{\sqrt x}\,y &= \frac{e^{-2\sqrt x}}{\sqrt x}\cdot e^{2\sqrt x} \\ \Rightarrow\ \frac{d}{dx}\left(y\,e^{2\sqrt x}\right) &= \frac{1}{\sqrt x}. \end{aligned}$$
   By the product and chain rules, $\frac{d}{dx}\left(y\,e^{2\sqrt x}\right) = e^{2\sqrt x}\frac{dy}{dx} + y\cdot\frac{e^{2\sqrt x}}{\sqrt x}$⁠, which is the left side. On the right, $e^{-2\sqrt x}\cdot e^{2\sqrt x} = e^0 = 1$⁠.
   Product rule (Chapter 5, Section 5.3). Basics: (y · g)′ = g(y′ + Py)

5. Integrate, with one constant

   $$\begin{aligned} y\,e^{2\sqrt x} &= \int\frac{e^{-2\sqrt x}}{\sqrt x}\cdot e^{2\sqrt x}\,dx + C \\ &= \int\frac{dx}{\sqrt x} + C = 2\sqrt x + C. \end{aligned}$$
   This is $y\,(\text{I.F.}) = \int(Q\times\text{I.F.})\,dx + C$⁠. The $C$ is the one arbitrary constant of this first order equation.
   Section 9.4.3, step (iii). Basics: one constant for order 1

6. The general solution

   $$\begin{aligned} &y\,e^{2\sqrt x} = 2\sqrt x + C, \\ &\text{that is, }\ y = (2\sqrt x + C)\,e^{-2\sqrt x}, \\ &\text{for } x > 0. \end{aligned}$$
   Multiply both sides by $e^{-2\sqrt x}$ to solve for $y$⁠. This is exactly the textbook's form $ye^{2\sqrt x} = (2\sqrt x + C)$⁠. Each value of $C$ gives one solution curve, defined on the whole interval $x > 0$⁠.
   Section 9.4.3: the general solution. Lesson: the Miscellaneous exercise player

7. Check by substituting back

   From $y = (2\sqrt x + C)e^{-2\sqrt x}$⁠, by the product and chain rules,

   $$\begin{aligned} \frac{dy}{dx} &= \frac{1}{\sqrt x}\,e^{-2\sqrt x} \\ &\qquad - (2\sqrt x + C)\,e^{-2\sqrt x}\cdot\frac{1}{\sqrt x} \\ &= \frac{e^{-2\sqrt x}}{\sqrt x} - \frac{y}{\sqrt x}, \end{aligned}$$

   which is the bracket of the given equation. So the bracket times $\frac{dx}{dy}$ is $\frac{dy}{dx}\cdot\frac{dx}{dy} = 1$⁠.

   $\frac{d}{dx}(2\sqrt x) = \frac{1}{\sqrt x}$ and $\frac{d}{dx}e^{-2\sqrt x} = -\frac{1}{\sqrt x}e^{-2\sqrt x}$⁠. Substituting the answer back returns the equation for every $C$ and every $x > 0$⁠, so it is the general solution.
   A solution makes L.H.S. = R.H.S. (Section 9.3). Lesson: general and particular solutions

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