Miscellaneous · Q2: Verify four solutions, from $xy = ae^x + be^{-x} + x^2$ to $x^2 = 2y^2\log y$

Basics you need

Part (i): $xy = ae^x + be^{-x} + x^2$

1. Differentiate the relation once

   Differentiate both sides of $xy = ae^x + be^{-x} + x^2\quad\ldots(1)$ with respect to $x$⁠. On the left, the product rule gives $x\frac{dy}{dx} + y$⁠:

   $$x\frac{dy}{dx} + y = ae^x - be^{-x} + 2x.\quad\ldots(2)$$
   The function is given implicitly ($y$ times $x$⁠), so we differentiate the relation as it stands rather than first dividing by $x$⁠. Both ways work for $x \ne 0$⁠; this one is shorter.
   Product rule; $(e^{-x})' = -e^{-x}$ (Chapter 5). Basics: y′ and y″

2. Differentiate again

   Differentiate (2): $x\frac{dy}{dx}$ gives $\frac{dy}{dx} + x\frac{d^2y}{dx^2}$ by the product rule, and $y$ gives $\frac{dy}{dx}$⁠:

   $$x\frac{d^2y}{dx^2} + 2\frac{dy}{dx} = ae^x + be^{-x} + 2.$$
   The equation to be verified is of second order, so we need $\frac{d^2y}{dx^2}$⁠. Notice that the sign of $be^{-x}$ has turned back to $+$⁠: differentiating $e^{-x}$ twice gives $e^{-x}$⁠.
   Second order derivative (Chapter 5, Section 5.7). Basics: y′ and y″

3. Remove $a$ and $b$ with (1), and conclude

   From (1), $ae^x + be^{-x} = xy - x^2$⁠. So

   $$\begin{aligned} &x\frac{d^2y}{dx^2} + 2\frac{dy}{dx} = xy - x^2 + 2, \\ &x\frac{d^2y}{dx^2} + 2\frac{dy}{dx} - xy \\ &\quad + x^2 - 2 = 0. \end{aligned}$$

   This is the given equation, for every $x \ne 0$ and all $a, b$⁠. So the function is a solution.

   The constants $a$ and $b$ appear in the same combination $ae^x + be^{-x}$ as in (1), so (1) replaces them by $xy - x^2$ at one stroke. What is left contains only $x$⁠, $y$ and the derivatives, which is what a differential equation may contain.
   Section 9.3: substitute and compare the two sides. Lesson: is it a solution?

Part (ii): $y = e^x(a\cos x + b\sin x)$

1. First derivative, written through $y$

   By the product rule,

   $$\begin{aligned} \frac{dy}{dx} &= e^x(a\cos x + b\sin x) \\ &\quad + e^x(-a\sin x + b\cos x) \\ &= y + e^x(b\cos x - a\sin x). \end{aligned}$$

   Call this (3).

   The first product, $e^x$ times the bracket, is $y$ itself, because $(e^x)' = e^x$⁠. Keeping it as $y$ makes the next step shorter.
   Product rule; $(\cos x)' = -\sin x$⁠, $(\sin x)' = \cos x$⁠. Basics: e^{kt} and its rate

2. Second derivative

   Differentiate (3). The derivative of $y$ is $\frac{dy}{dx}$⁠; the product $e^x(b\cos x - a\sin x)$ gives two terms:

   $$\begin{aligned} \frac{d^2y}{dx^2} &= \frac{dy}{dx} + e^x(b\cos x - a\sin x) \\ &\quad + e^x(-b\sin x - a\cos x). \end{aligned}$$

   By (3), $e^x(b\cos x - a\sin x) = \frac{dy}{dx} - y$⁠. And the last term is $-e^x(a\cos x + b\sin x)$⁠, which is $-y$⁠. So

   $$\begin{aligned} \frac{d^2y}{dx^2} &= \frac{dy}{dx} + \left(\frac{dy}{dx} - y\right) - y \\ &= 2\frac{dy}{dx} - 2y. \end{aligned}$$
   Every piece is now written through $y$ and $\frac{dy}{dx}$⁠, so $a$ and $b$ have disappeared.
   Differentiate (3) term by term (Chapter 5). Basics: y′ and y″

3. Substitute and conclude

   $$\begin{aligned} &\frac{d^2y}{dx^2} - 2\frac{dy}{dx} + 2y \\ &= \left(2\frac{dy}{dx} - 2y\right) - 2\frac{dy}{dx} + 2y = 0. \end{aligned}$$

   So L.H.S. = 0 = R.H.S. for every real $x$ and all $a, b$⁠: the function is a solution, with two arbitrary constants for a second order equation.

   This is a special case of Example 19, where $e^{ax}(c_1\cos bx + c_2\sin bx)$ solves $y'' - 2ay' + (a^2 + b^2)y = 0$⁠. Taking the exponent and the frequency there both equal to 1 gives our function and our equation $y'' - 2y' + 2y = 0$⁠.
   Section 9.3: L.H.S. = R.H.S. Lesson: Examples 19 to 22

Part (iii): $y = x\sin 3x$

1. First derivative

   $$\begin{aligned} \frac{dy}{dx} &= \sin 3x + x\cdot 3\cos 3x \\ &= \sin 3x + 3x\cos 3x. \end{aligned}$$
   $y$ is a product of $x$ and $\sin 3x$⁠; the chain rule gives $(\sin 3x)' = 3\cos 3x$⁠.
   Product rule and chain rule (Chapter 5). Basics: y′ and y″

2. Second derivative

   $$\begin{aligned} \frac{d^2y}{dx^2} &= 3\cos 3x + \left(3\cos 3x - 9x\sin 3x\right) \\ &= 6\cos 3x - 9x\sin 3x. \end{aligned}$$
   The first term, $\sin 3x$⁠, gives $3\cos 3x$⁠. The second, $3x\cos 3x$⁠, is again a product: $3\cos 3x + 3x(-3\sin 3x)$⁠.
   $(\cos 3x)' = -3\sin 3x$ (Chapter 5). Basics: y′ and y″

3. Substitute and conclude

   $$\begin{aligned} &\frac{d^2y}{dx^2} + 9y - 6\cos 3x \\ &= 6\cos 3x - 9x\sin 3x \\ &\quad + 9x\sin 3x - 6\cos 3x \\ &= 0. \end{aligned}$$

   So L.H.S. = 0 = R.H.S. for every real $x$⁠: the function is a solution. It has no arbitrary constant, so it is a particular solution of this second order equation.

   The $9x\sin 3x$ terms cancel, and so do the $6\cos 3x$ terms. The first derivative was only a stage on the way: $\frac{dy}{dx}$ does not appear in the equation.
   Section 9.3: a solution with no arbitrary constant is a particular solution. Basics: order n gives n constants

Part (iv): $x^2 = 2y^2\log y$

1. Where the relation lives

   $\log y$ needs $y > 0$⁠. The left side $x^2$ is never negative, so $2y^2\log y \ge 0$⁠, that is $\log y \ge 0$⁠, that is $y \ge 1$⁠. Hence

   $$1 + 2\log y \ge 1 > 0.$$
   We will divide by $1 + 2\log y$ in the next step, so we first make sure it is never 0. On this curve it cannot be: $1 + 2\log y = 0$ would need $y = e^{-1/2} < 1$⁠.
   $\log y$ is defined for $y > 0$ and is $\ge 0$ exactly when $y \ge 1$ (Chapter 5, Section 5.4). Basics: e^{kt} and log

2. Differentiate implicitly and find $\frac{dy}{dx}$

   Differentiate both sides with respect to $x$⁠, using the product rule on $y^2\cdot\log y$⁠:

   $$\begin{aligned} 2x &= 2\left(2y\log y + y^2\cdot\frac{1}{y}\right)\frac{dy}{dx} \\ &= 2y(1 + 2\log y)\frac{dy}{dx}, \\ \frac{dy}{dx} &= \frac{x}{y(1 + 2\log y)}. \end{aligned}$$
   Each factor that contains $y$ carries a $\frac{dy}{dx}$ by the chain rule. The divisor $y(1 + 2\log y)$ is positive by Part (iv), Step 1, so the division is allowed.
   Implicit differentiation; $(\log y)' = \frac{1}{y}\frac{dy}{dx}$ (Chapter 5). Basics: y′ and y″

3. Use the relation to rewrite $x^2 + y^2$⁠, substitute and conclude

   By the given relation, $x^2 + y^2 = 2y^2\log y + y^2 = y^2(1 + 2\log y)$⁠. So

   $$\begin{aligned} &(x^2 + y^2)\frac{dy}{dx} - xy \\ &= y^2(1 + 2\log y)\cdot\frac{x}{y(1 + 2\log y)} - xy \\ &= xy - xy = 0. \end{aligned}$$

   So L.H.S. = 0 = R.H.S. at every point of the curve: the relation is a solution of the equation.

   The factor $1 + 2\log y$ appears in both places and cancels, and $\frac{y^2}{y} = y$⁠. The relation was used twice: once to differentiate, once to replace $x^2$⁠.
   Section 9.3: an implicit relation is a solution when it makes L.H.S. = R.H.S. Lesson: Miscellaneous exercise player

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