Miscellaneous · Q14 (MCQ): General solution of $\frac{dx}{dy} + P_1x = Q_1$

Basics you need

Solution

1. Which variable is the unknown?

   The equation gives $\frac{dx}{dy}$⁠, so $x$ is the unknown function of $y$⁠. It has the form $\frac{dx}{dy} + P_1x = Q_1$ with $P_1$⁠, $Q_1$ depending on $y$ only: linear in $x$⁠.

   $x$ and $\frac{dx}{dy}$ appear only to the first power and are not multiplied together. This is the "another form" of a first order linear equation in Section 9.4.3: the form $\frac{dy}{dx} + Py = Q$ with $x$ and $y$ swapped. Order 1, degree 1.
   Section 9.4.3: the form $\frac{dx}{dy} + P_1x = Q_1$⁠. Lesson: linear equations

2. The integrating factor: $g(y) = e^{\int P_1\,dy}$

   Multiply the equation by $g(y) = e^{\int P_1\,dy}$⁠. Since $g'(y) = P_1\,g(y)$⁠,

   $$g\frac{dx}{dy} + P_1g\,x = g\frac{dx}{dy} + g'x = \frac{d}{dy}(x\,g),$$

   so the equation becomes $\frac{d}{dy}\left(x\,e^{\int P_1\,dy}\right) = Q_1e^{\int P_1\,dy}$⁠.

   By the chain rule, $\frac{d}{dy}e^{\int P_1\,dy} = e^{\int P_1\,dy}\cdot P_1$⁠. The product rule with respect to $y$ then turns the left side into one derivative. This is the book's derivation in Section 9.4.3, with $x$ and $y$ swapped.
   Section 9.4.3, the integrating factor. Basics: (y · g)′ = g(y′ + Py)

3. Integrate with respect to $y$

   $$x\,e^{\int P_1\,dy} = \int\left(Q_1e^{\int P_1\,dy}\right)dy + C.$$
   Both sides are functions of $y$⁠, so we integrate with respect to $y$⁠, adding one arbitrary constant $C$⁠. This is $x\cdot(\text{I.F.}) = \int(Q_1\times\text{I.F.})\,dy + C$⁠, the solution the book states for this form.
   Section 9.4.3. Basics: one constant for order 1

4. Choose the option

   The general solution is $x\,e^{\int P_1\,dy} = \int\left(Q_1e^{\int P_1\,dy}\right)dy + C$⁠. The correct option is (C).

   It multiplies the unknown $x$ by the I.F. $e^{\int P_1\,dy}$ and integrates $dy$⁠, exactly as derived.
   The multiple-choice questions of the chapter. Lesson: the multiple-choice questions

5. Examine every option

   Option	What it does	Verdict
   (A)	Right I.F. and $dy$⁠, but multiplies $y$⁠, not the unknown $x$	Wrong
   (B)	The rule for $\frac{dy}{dx} + Py = Q$⁠: multiplies $y$⁠, integrates $dx$	Wrong form
   (C)	Multiplies $x$ by $e^{\int P_1\,dy}$⁠, integrates $dy$	Correct
   (D)	Multiplies $x$⁠, but integrates $P_1$⁠, $Q_1$ (functions of $y$⁠) with respect to $x$	Wrong

   Test them on $\frac{dx}{dy} + x = 1$ ($P_1 = Q_1 = 1$⁠), whose solutions are $x = 1 + Ce^{-y}$⁠:

   (C): $x\,e^{y} = \int e^{y}\,dy + C = e^{y} + C$⁠, so $x = 1 + Ce^{-y}$⁠. Correct.

   (A): $y\,e^{y} = e^{y} + C$ makes $y$ a constant: horizontal lines, along which $\frac{dx}{dy}$ does not exist.

   (B): $y\,e^{x} = e^{x} + C$⁠, so $y = 1 + Ce^{-x}$⁠. With $C = 1$⁠, at $(0, 2)$⁠: $\frac{dy}{dx} = -1$⁠, so $\frac{dx}{dy} + x = -1 + 0 = -1 \ne 1$⁠.

   (D): $x\,e^{x} = e^{x} + C$ makes $x$ a constant $k$⁠: then $\frac{dx}{dy} = 0$ and the equation needs $k = 1$⁠, a single line, not a family with an arbitrary constant.

   The test equation shows each failure concretely: only (C) produces the whole family $x = 1 + Ce^{-y}$⁠. In general, $P_1$ and $Q_1$ are functions of $y$⁠, so integrals $dx$ of them, as in (B) and (D), are the wrong integrals, and the I.F. must multiply the unknown $x$⁠, not $y$⁠.
   Section 9.4.3 and its examples in $x$⁠. Compare Example 16. Lesson: Examples 14 to 18

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