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Chapter 9 · Differential Equations

Miscellaneous · Q15 (MCQ): General solution of ex dy+(yex+2x) dx=0e^x\,dy + (ye^x + 2x)\,dx = 0

Miscellaneous ExerciseLinear differential equations · Section 9.4.3
The question

15. The general solution of the differential equation ex dy+(y ex+2x) dx=0e^x\,dy + (y\,e^x + 2x)\,dx = 0 is

(A) x ey+x2=Cx\,e^y + x^2 = \text{C}    (B) x ey+y2=Cx\,e^y + y^2 = \text{C}

(C) y ex+x2=Cy\,e^x + x^2 = \text{C}    (D) y ey+x2=Cy\,e^y + x^2 = \text{C}

Exam answer

Dividing by ex dxe^x\,dx⁠: dydx+y=−2xe−x\frac{dy}{dx} + y = -2xe^{-x}⁠, a linear differential equation with P=1P = 1⁠, Q=−2xe−xQ = -2xe^{-x}⁠.

I.F.=e∫1 dx=ex\text{I.F.} = e^{\int 1\,dx} = e^x⁠, so y ex=∫−2x dx+Cy\,e^x = \int -2x\,dx + C =−x2+C= -x^2 + C⁠.

Answer: (C) y ex+x2=Cy\,e^x + x^2 = C

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

Divided by ex dxe^x\,dx⁠, the equation is linear: dydx+y=−2xe−x\frac{dy}{dx} + y = -2xe^{-x}⁠, with I.F. e∫1 dx=exe^{\int 1\,dx} = e^x⁠. Multiplying back by exe^x returns the given left side exdydx+yexe^x\frac{dy}{dx} + ye^x⁠, which is ddx(yex)\frac{d}{dx}(ye^x)⁠. So the equation says ddx(yex+x2)=0\frac{d}{dx}(ye^x + x^2) = 0⁠, and yex+x2=Cye^x + x^2 = C⁠. Option (C).

Given, to find, and the plan
Given
ex dy+(yex+2x) dx=0e^x\,dy + (ye^x + 2x)\,dx = 0⁠.
To find
Its general solution, and which option gives it.
Plan
(1) Write it in the linear form; read off PP and QQ⁠. (2) I.F. =e∫P dx= e^{\int P\,dx}⁠. (3) Integrate, with one constant. (4) Check by substituting back. (5) Choose the option. (6) Examine every option.

Basics you need

1First order linear equations

dydx+Py=Q\frac{dy}{dx} + Py = Q⁠; I.F. =e∫P dx= e^{\int P\,dx}⁠; y (I.F.)=∫(Q×I.F.) dx+Cy\,(\text{I.F.}) = \int(Q\times\text{I.F.})\,dx + C (Section 9.4.3). Lesson: linear equations, the integrating factor

2The product rule

ddx(yex)=exdydx+yex\frac{d}{dx}(ye^x) = e^x\frac{dy}{dx} + ye^x⁠. Basics: (y · g)′ = g(y′ + Py)

3Implicit differentiation

To find the slope of a curve G(x,y)=CG(x, y) = C⁠, differentiate both sides with respect to xx⁠, treating yy as a function of xx (Chapter 5, Section 5.3.2). Basics: what a derivative means

Solution

  1. Write it in the linear form

    Divide by ex dxe^x\,dx (ex>0e^x > 0⁠):

    dydx+y+2xe−x=0⇒ dydx+y=−2xe−x.\begin{aligned} \frac{dy}{dx} + y + 2xe^{-x} &= 0 \\ \Rightarrow\ \frac{dy}{dx} + y &= -2xe^{-x}. \end{aligned}
    yexex=y\frac{ye^x}{e^x} = y and 2xex=2xe−x\frac{2x}{e^x} = 2xe^{-x}⁠. Now P=1P = 1 and Q=−2xe−xQ = -2xe^{-x} depend on xx only (PP is a constant), so the equation is linear, defined for every real xx⁠. Order 1, degree 1.
    Section 9.4.3, step (i); Sections 9.2.1 and 9.2.2. Lesson: linear equations

  2. The integrating factor: exe^x

    I.F.=e∫1 dx=ex.\text{I.F.} = e^{\int 1\,dx} = e^x.
    ∫1 dx=x\int 1\,dx = x⁠. Multiplying the linear form by exe^x gives exdydx+yex=−2xe^x\frac{dy}{dx} + ye^x = -2x⁠, whose left side is ddx(yex)\frac{d}{dx}(ye^x) by the product rule; it is the original equation again, divided by dxdx⁠.
    Section 9.4.3, step (ii). Basics: (y · g)′ = g(y′ + Py)

  3. Integrate, with one constant

    y ex=∫(−2xe−x) ex dx+C=∫−2x dx+C=−x2+C,\begin{aligned} y\,e^x &= \int(-2xe^{-x})\,e^x\,dx + C \\ &= \int -2x\,dx + C = -x^2 + C, \end{aligned}

    that is,

    y ex+x2=C.y\,e^x + x^2 = C.
    This is y (I.F.)=∫(Q×I.F.) dx+Cy\,(\text{I.F.}) = \int(Q\times\text{I.F.})\,dx + C⁠; e−xex=1e^{-x}e^x = 1⁠, and ∫−2x dx=−x2\int -2x\,dx = -x^2⁠. Moving −x2-x^2 across gives the form of the options. The CC is the one arbitrary constant.
    Section 9.4.3, step (iii); Section 7.2, formula (i). Basics: one constant for order 1

  4. Check by substituting back

    Differentiate y ex+x2=Cy\,e^x + x^2 = C with respect to xx⁠:

    exdydx+y ex+2x=0,e^x\frac{dy}{dx} + y\,e^x + 2x = 0,

    and multiplying by dxdx gives ex dy+(yex+2x) dx=0e^x\,dy + (ye^x + 2x)\,dx = 0⁠, the given equation.

    The product rule differentiates yexye^x⁠; the constant CC has derivative 0. Substituting the answer back returns the equation for every CC⁠, and there is one constant for this first order equation, so it is the general solution.
    A solution makes L.H.S. = R.H.S. (Section 9.3). Lesson: general and particular solutions

  5. Choose the option

    The general solution is y ex+x2=Cy\,e^x + x^2 = C⁠. The correct option is (C).

    It is exactly option (C).
    The multiple-choice questions of the chapter. Lesson: the multiple-choice questions

  6. Examine every option

    At the point (1,0)(1, 0) the equation asks for the slope dydx=−yex+2xex=−2e\frac{dy}{dx} = -\frac{ye^x + 2x}{e^x} = -\frac{2}{e}⁠. Differentiate each option implicitly and evaluate at (1,0)(1, 0)⁠:

    OptionSlope of the familyAt (1,0)(1, 0)
    (A) xey+x2=Cxe^y + x^2 = Cdydx=−ey+2xxey\frac{dy}{dx} = -\frac{e^y + 2x}{xe^y}−3-3⁠: wrong
    (B) xey+y2=Cxe^y + y^2 = Cdydx=−eyxey+2y\frac{dy}{dx} = -\frac{e^y}{xe^y + 2y}−1-1⁠: wrong
    (C) yex+x2=Cye^x + x^2 = Cdydx=−yex+2xex\frac{dy}{dx} = -\frac{ye^x + 2x}{e^x}−2e-\frac{2}{e}⁠: correct
    (D) yey+x2=Cye^y + x^2 = Cdydx=−2xey(1+y)\frac{dy}{dx} = -\frac{2x}{e^y(1 + y)}−2-2⁠: wrong
    For (A): ey+xeydydx+2x=0e^y + xe^y\frac{dy}{dx} + 2x = 0⁠. For (B): ey+xeydydx+2ydydx=0e^y + xe^y\frac{dy}{dx} + 2y\frac{dy}{dx} = 0⁠. For (D): (ey+yey)dydx+2x=0(e^y + ye^y)\frac{dy}{dx} + 2x = 0⁠. A family is the general solution only if its slope equals the equation's slope at every point; one point where they differ rules an option out.
    Implicit differentiation (Chapter 5, Section 5.3.2). Lesson: Examples 14 to 18
Result

The general solution of ex dy+(yex+2x) dx=0e^x\,dy + (ye^x + 2x)\,dx = 0 is y ex+x2=Cy\,e^x + x^2 = C⁠: option (C).

(Textbook answer: C.)

Check it with numbers

At (1,0)(1, 0) the equation's slope is −2e≈−0.7357589-\frac{2}{e} \approx -0.7357589⁠; the options give (A) −3-3⁠, (B) −1-1⁠, (C) −0.7357589-0.7357589⁠, (D) −2-2⁠. The member of (C) through (1,0)(1, 0) is y=(1−x2)e−xy = (1 - x^2)e^{-x} (C=1C = 1⁠), and its slope is [−2x−(1−x2)]e−x\left[-2x - (1 - x^2)\right]e^{-x}⁠, which at x=1x = 1 is −2e−1≈−0.7357589-2e^{-1} \approx -0.7357589⁠.

Common mistakes
  1. Dividing by exe^x but not 2x2x⁠: Q=−2xex=−2xe−xQ = -\frac{2x}{e^x} = -2xe^{-x}⁠, not −2x-2x⁠.
  2. Mixing up exe^x and eye^y in the options: only (C) has yexye^x⁠, the term that comes from ex dy+yex dxe^x\,dy + ye^x\,dx⁠.
What to take away
  1. ex dy+yex dx=d(yex)e^x\,dy + ye^x\,dx = d(ye^x)⁠: spotting a product rule in the equation can solve it in one line.
  2. To test an option, differentiate it implicitly and compare slopes at one convenient point.

Every fact used, and where it comes from

FactUsed inSource
Linear equation; I.F. =e∫P dx= e^{\int P\,dx}⁠; y (I.F.)=∫(Q×I.F.) dx+Cy\,(\text{I.F.}) = \int(Q\times\text{I.F.})\,dx + CSteps 1, 2, 3Section 9.4.3, steps (i) to (iii) (Lesson)
Order and degreeStep 1Sections 9.2.1 and 9.2.2 (Lesson)
∫x dx=x22+C\int x\,dx = \frac{x^2}{2} + CStep 3Section 7.2, formula (i) (Basics)
Product rule; implicit differentiationSteps 2, 4, 6Chapter 5, Sections 5.3 and 5.3.2 (Basics)
Substituting a solution backStep 4Section 9.3 (Lesson)
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