Class 12 Maths, Worked in Full Past papersGet updates
Chapter 9 · Differential Equations

Exercise 9.3 · Q23 (MCQ): General solution of dydx=ex+y\frac{dy}{dx} = e^{x + y}

Exercise 9.3Variables separable · Section 9.4.1
The question

23. The general solution of the differential equation dydx=ex+y\dfrac{dy}{dx} = e^{x + y} is

(A) ex+e−y=Ce^x + e^{-y} = C    (B) ex+ey=Ce^x + e^y = C    (C) e−x+ey=Ce^{-x} + e^y = C    (D) e−x+e−y=Ce^{-x} + e^{-y} = C

Exam answer

dydx=exey⇒e−y dy=ex dx\frac{dy}{dx} = e^xe^y \Rightarrow e^{-y}\,dy = e^x\,dx⁠.

Integrating, −e−y=ex+K-e^{-y} = e^x + K⁠, so ex+e−y=Ce^x + e^{-y} = C⁠, where C=−KC = -K⁠.

Answer: (A) ex+e−y=Ce^x + e^{-y} = C

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

The one idea this problem is built on

ex+y=ex⋅eye^{x + y} = e^x\cdot e^y⁠, a function of xx times a function of yy⁠, so the equation is separable. Dividing by eye^y (multiplying by e−ye^{-y}⁠) and integrating gives −e−y=ex+K-e^{-y} = e^x + K⁠; moving terms gives option (A).

Given, to find, and the plan
Given
dydx=ex+y\frac{dy}{dx} = e^{x + y} and four candidate general solutions.
To find
The option that is the general solution.
Plan
(1) Split the exponential. (2) Separate. (3) Integrate, one constant. (4) Rearrange and choose the option. (5) Examine every option by differentiating it.

Basics you need

1Index law

ex+y=exeye^{x + y} = e^xe^y and 1ey=e−y\frac{1}{e^y} = e^{-y}⁠.

2Variables separable

If dydx=g(x) h(y)\frac{dy}{dx} = g(x)\,h(y) with h(y)≠0h(y) \ne 0⁠, then ∫dyh(y)=∫g(x) dx\int\frac{dy}{h(y)} = \int g(x)\,dx (Section 9.4.1). Lesson: separate, then integrate

3Exponential integrals

∫ex dx=ex+C\int e^x\,dx = e^x + C and ∫e−y dy=−e−y+C\int e^{-y}\,dy = -e^{-y} + C (Section 7.2). Basics: every method ends with an integral

4Checking a general solution

Differentiate the relation implicitly; a solution gives back the equation (Section 9.3). Basics: one constant, one condition

Solution

  1. Split the exponential and separate

    dydx=exey  ⇒  e−y dy=ex dx.\frac{dy}{dx} = e^xe^y \;\Rightarrow\; e^{-y}\,dy = e^x\,dx.
    The index law makes the right side g(x) h(y)g(x)\,h(y) with g(x)=exg(x) = e^x⁠, h(y)=eyh(y) = e^y⁠. Dividing by eye^y is always allowed, since ey>0e^y > 0⁠, so no solution is lost. The equation has order 1 and degree 1.
    Section 9.4.1: separate, then integrate. Lesson: separate, then integrate

  2. Integrate both sides, with one constant

    ∫e−y dy=∫ex dx⇒ −e−y=ex+K.\begin{aligned} \int e^{-y}\,dy &= \int e^x\,dx \\ \Rightarrow\ -e^{-y} &= e^x + K. \end{aligned}
    The derivative of −e−y-e^{-y} with respect to yy is e−ye^{-y}⁠, so ∫e−y dy=−e−y\int e^{-y}\,dy = -e^{-y}⁠. The constants of both sides combine into one, KK⁠.
    Section 7.2. Basics: every method ends with an integral

  3. Rearrange and choose the option

    Add e−ye^{-y} and subtract KK on both sides:

    ex+e−y=−K=C.e^x + e^{-y} = -K = C.

    The correct option is (A).

    C=−KC = -K is again an arbitrary constant, so this is the general solution in the form of option (A). Since ex+e−y>0e^x + e^{-y} > 0⁠, only positive values of CC actually give curves.
    Renaming the constant. Basics: one constant, one condition

  4. Examine every option

    Differentiate each relation with respect to xx and solve for dydx\frac{dy}{dx}⁠:

    OptionDifferentiating givesVerdict
    (A) ex+e−y=Ce^x + e^{-y} = Cex−e−yy′=0e^x - e^{-y}y' = 0⁠, so y′=ex+yy' = e^{x + y}Correct
    (B) ex+ey=Ce^x + e^y = Cex+eyy′=0e^x + e^yy' = 0⁠, so y′=−ex−yy' = -e^{x - y}Wrong: sign and exponent
    (C) e−x+ey=Ce^{-x} + e^y = C−e−x+eyy′=0-e^{-x} + e^yy' = 0⁠, so y′=e−x−yy' = e^{-x - y}Wrong: exponent −(x+y)-(x + y)
    (D) e−x+e−y=Ce^{-x} + e^{-y} = C−e−x−e−yy′=0-e^{-x} - e^{-y}y' = 0⁠, so y′=−ey−xy' = -e^{y - x}Wrong: sign and exponent
    Only (A) gives back dydx=ex+y\frac{dy}{dx} = e^{x + y}⁠. Each of (B), (C) and (D) differs from (A) by the sign of one or both exponents, and each sign change changes the slope the relation gives, so none of them is a solution.
    Implicit differentiation (Chapter 5). Basics: y′ is the slope
Result

The general solution is ex+e−y=Ce^x + e^{-y} = C⁠: option (A).

(Textbook answer: A.)

Check it with numbers

Take C=2C = 2⁠: the curve ex+e−y=2e^x + e^{-y} = 2 passes through (0,0)(0, 0)⁠. Solving for yy⁠: y=−log⁡(2−ex)y = -\log(2 - e^x) for x<log⁡2x < \log 2⁠.

At x=0x = 0 the equation gives the slope e0+0=1e^{0 + 0} = 1⁠. From the curve: y(0.001)≈0.0010010y(0.001) \approx 0.0010010 and y(−0.001)≈−0.0009990y(-0.001) \approx -0.0009990⁠, so

0.0010010+0.00099900.002≈1.0000.\begin{aligned} &\frac{0.0010010 + 0.0009990}{0.002} \\ &\approx 1.0000. \end{aligned}
Common mistakes
  1. Writing ex+y=ex+eye^{x + y} = e^x + e^y⁠. The index law turns a sum in the exponent into a product.
  2. ∫e−y dy=e−y\int e^{-y}\,dy = e^{-y}⁠. The minus sign in the exponent brings a minus sign outside: −e−y-e^{-y}⁠.
What to take away
  1. ex+y=exeye^{x + y} = e^xe^y makes the equation separable.
  2. In an MCQ, differentiating each option is a quick, sure check.

Every fact used, and where it comes from

FactUsed inSource
ex+y=exeye^{x + y} = e^xe^y⁠; variables separableStep 1Index laws; Section 9.4.1 (Lesson)
∫ex dx=ex+C\int e^x\,dx = e^x + C⁠; ∫e−y dy=−e−y+C\int e^{-y}\,dy = -e^{-y} + CStep 2Section 7.2 (Basics)
An arbitrary constant may be renamed (C=−KC = -K⁠)Step 3Section 9.3 (Basics)
Implicit differentiationStep 4Chapter 5 (Basics)
Found an error, or want more?

Tell us if something on this page is wrong or unclear, or if you would like a fuller explanation, another method or a new feature. Send feedback on this problem →