Calculus 1 Quiz: General Solutions Separation Of Variables
20 questions · exam conditions
0:00
General Solutions Separation Of VariablesQuestion 1 of 20

What is the general solution to xdydx=yln⁡(x)x \frac{dy}{dx} = y \ln(x)?

y=C+12(ln⁡x)2y = C + \frac{1}{2}(\ln x)^2
y=Cxln⁡xy = Cx^{\ln x}
y=Ce12(ln⁡x)2y = C e^{\frac{1}{2}(\ln x)^2}
y=Cln⁡(x2)y = C \ln(x^2)
← Back to quizzes

Calculus 1 Quiz

Calculus 1 Quiz: General Solutions Separation Of Variables

Practice General Solutions Separation Of Variables in Calculus 1 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on General Solutions Separation Of Variables, giving you a quick way to practice the rules, question types, and explanations that matter most for Calculus 1.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

What is the general solution to xdydx=yln⁡(x)x \frac{dy}{dx} = y \ln(x)?

  1. y=C+12(ln⁡x)2y = C + \frac{1}{2}(\ln x)^2
  2. y=Cxln⁡xy = Cx^{\ln x}
  3. y=Ce12(ln⁡x)2y = C e^{\frac{1}{2}(\ln x)^2} (correct answer)
  4. y=Cln⁡(x2)y = C \ln(x^2)
Explanation: Separate the variables: 1ydy=ln⁡(x)xdx\frac{1}{y}dy = \frac{\ln(x)}{x}dx. Integrate both sides. For the right side, use u-substitution with u=ln⁡(x)u = \ln(x), so du=1xdxdu = \frac{1}{x}dx. The integral becomes ∫1ydy=∫udu\int \frac{1}{y}dy = \int u du. This gives ln⁡∣y∣=12u2+K=12(ln⁡x)2+K\ln|y| = \frac{1}{2}u^2 + K = \frac{1}{2}(\ln x)^2 + K. To solve for y, exponentiate both sides: ∣y∣=e12(ln⁡x)2+K=eKe12(ln⁡x)2|y| = e^{\frac{1}{2}(\ln x)^2 + K} = e^K e^{\frac{1}{2}(\ln x)^2}. Let C=±eKC = \pm e^K. The solution is y=Ce12(ln⁡x)2y = C e^{\frac{1}{2}(\ln x)^2}.

Question 2

Which of the following represents the general solution to the differential equation y′=e2x+yy' = e^{2x+y}?

  1. y=ln⁡(12e2x+C)y = \ln(\frac{1}{2}e^{2x} + C)
  2. y=−ln⁡(−12e2x+C)y = -\ln(-\frac{1}{2}e^{2x} + C) (correct answer)
  3. y=12e2x+ey+Cy = \frac{1}{2}e^{2x} + e^y + C
  4. y=−2e−2x+Cy = -2e^{-2x} + C
Explanation: First, use exponent rules to write y′=e2xeyy' = e^{2x}e^y. Separate the variables: e−ydy=e2xdxe^{-y}dy = e^{2x}dx. Integrate both sides: ∫e−ydy=∫e2xdx\int e^{-y}dy = \int e^{2x}dx, which gives −e−y=12e2x+K-e^{-y} = \frac{1}{2}e^{2x} + K. To solve for y, first multiply by -1: e−y=−12e2x−Ke^{-y} = -\frac{1}{2}e^{2x} - K. Let C=−KC = -K, so e−y=C−12e2xe^{-y} = C - \frac{1}{2}e^{2x}. Now, take the natural logarithm of both sides: −y=ln⁡(C−12e2x)-y = \ln(C - \frac{1}{2}e^{2x}). Finally, y=−ln⁡(C−12e2x)y = -\ln(C - \frac{1}{2}e^{2x}), which is equivalent to choice B.

Question 3

The general solution to the differential equation dydx=−4x(y−1)\frac{dy}{dx} = -4x(y-1) is of the form y=1+f(x,C)y = 1 + f(x, C). What is f(x,C)f(x, C)?

  1. C(y−1)e−2x2C(y-1)e^{-2x^2}
  2. Ce−4xCe^{-4x}
  3. e−2x2+Ce^{-2x^2} + C
  4. Ce−2x2Ce^{-2x^2} (correct answer)
Explanation: When you encounter a differential equation like dydx=−4x(y−1)\frac{dy}{dx} = -4x(y-1), recognize this as a separable differential equation. The key insight is that you can separate variables by moving all terms involving yy to one side and all terms involving xx to the other. Starting with dydx=−4x(y−1)\frac{dy}{dx} = -4x(y-1), separate variables: dyy−1=−4x dx\frac{dy}{y-1} = -4x \, dx Now integrate both sides. The left side gives ln⁡∣y−1∣\ln|y-1|, and the right side gives −2x2+C1-2x^2 + C_1. So: ln⁡∣y−1∣=−2x2+C1\ln|y-1| = -2x^2 + C_1 Exponentiating both sides: ∣y−1∣=e−2x2+C1=eC1⋅e−2x2|y-1| = e^{-2x^2 + C_1} = e^{C_1} \cdot e^{-2x^2} Since eC1e^{C_1} is just another constant, call it CC: y−1=Ce−2x2y-1 = Ce^{-2x^2} Therefore: y=1+Ce−2x2y = 1 + Ce^{-2x^2} This matches the form y=1+f(x,C)y = 1 + f(x,C) where f(x,C)=Ce−2x2f(x,C) = Ce^{-2x^2}, which is choice D. Choice A incorrectly includes (y−1)(y-1) in the function ff, but ff should only depend on xx and CC. Choice B has the wrong exponent—it uses −4x-4x instead of −2x2-2x^2, missing the integration of xx. Choice C adds CC to the exponential instead of multiplying by it, which doesn't arise from the integration process. Study tip: For separable differential equations, always check your integration carefully—especially when integrating polynomials like xx, where ∫−4x dx=−2x2\int -4x \, dx = -2x^2, not −4x-4x.

Question 4

Find the general solution to the differential equation y′=y+exyy' = y + e^x y.

  1. y=C(x+ex)y = C(x+e^x)
  2. y=ex+ex+Cy = e^{x+e^x} + C
  3. y=Cex+exy = C e^{x+e^x} (correct answer)
  4. y=Cex+Ceexy = C e^x + C e^{e^x}
Explanation: First, factor the right side: y′=y(1+ex)y' = y(1+e^x). Separate the variables: 1ydy=(1+ex)dx\frac{1}{y}dy = (1+e^x)dx. Integrate both sides: ∫1ydy=∫(1+ex)dx\int \frac{1}{y}dy = \int (1+e^x)dx. This gives ln⁡∣y∣=x+ex+K\ln|y| = x + e^x + K. Exponentiate both sides: ∣y∣=ex+ex+K=eKex+ex|y| = e^{x+e^x+K} = e^K e^{x+e^x}. Let C=±eKC = \pm e^K. The general solution is y=Cex+exy = C e^{x+e^x}.

Question 5

What is the implicit general solution to the differential equation (1+y2)dx+(1+x2)dy=0(1+y^2)dx + (1+x^2)dy = 0?

  1. arctan⁡(y)+arctan⁡(x)=C\arctan(y) + \arctan(x) = C (correct answer)
  2. ln⁡(1+y2)+ln⁡(1+x2)=C\ln(1+y^2) + \ln(1+x^2) = C
  3. y+y3/3+x+x3/3=Cy + y^3/3 + x + x^3/3 = C
  4. arctan⁡(y)−arctan⁡(x)=C\arctan(y) - \arctan(x) = C
Explanation: Rearrange the equation to (1+x2)dy=−(1+y2)dx(1+x^2)dy = -(1+y^2)dx. Separate variables: dy1+y2=−dx1+x2\frac{dy}{1+y^2} = -\frac{dx}{1+x^2}. Integrate both sides: ∫dy1+y2=−∫dx1+x2\int \frac{dy}{1+y^2} = -\int \frac{dx}{1+x^2}. This gives arctan⁡(y)=−arctan⁡(x)+C\arctan(y) = -\arctan(x) + C. Rearranging gives the implicit solution arctan⁡(y)+arctan⁡(x)=C\arctan(y) + \arctan(x) = C.

Question 6

Determine the general solution of the differential equation dydx−x2y=0\frac{dy}{dx} - x^2y = 0.

  1. y=Cex3/3y = C e^{x^3/3} (correct answer)
  2. y=ex3/3+Cy = e^{x^3/3} + C
  3. y=Cx3/3y = Cx^3/3
  4. y=x2+Cy = x^2 + C
Explanation: First, rearrange the equation to dydx=x2y\frac{dy}{dx} = x^2y. Separate the variables: 1ydy=x2dx\frac{1}{y}dy = x^2 dx. Integrate both sides: ∫1ydy=∫x2dx\int \frac{1}{y}dy = \int x^2 dx, which results in ln⁡∣y∣=x33+K\ln|y| = \frac{x^3}{3} + K. Exponentiating both sides gives ∣y∣=ex3/3+K=eKex3/3|y| = e^{x^3/3 + K} = e^K e^{x^3/3}. Let C=±eKC = \pm e^K. The general solution is y=Cex3/3y = C e^{x^3/3}.

Question 7

Find the general solution for y(x)y(x) given that y′=x1−y2y' = x\sqrt{1-y^2}.

  1. y(x)=sin⁡(x22)+Cy(x) = \sin(\frac{x^2}{2}) + C
  2. y(x)=cos⁡(x22+C)y(x) = \cos(\frac{x^2}{2} + C)
  3. y(x)=sin⁡(x22+C)y(x) = \sin(\frac{x^2}{2} + C) (correct answer)
  4. y(x)=arctan⁡(x22+C)y(x) = \arctan(\frac{x^2}{2} + C)
Explanation: Separate the variables: 11−y2dy=x dx\frac{1}{\sqrt{1-y^2}}dy = x\,dx. Integrate both sides: ∫11−y2dy=∫x dx\int \frac{1}{\sqrt{1-y^2}}dy = \int x\,dx. The left integral is arcsin⁡(y)\arcsin(y) and the right is x22+C\frac{x^2}{2} + C. So, arcsin⁡(y)=x22+C\arcsin(y) = \frac{x^2}{2} + C. To solve for yy, take the sine of both sides: y=sin⁡(x22+C)y = \sin(\frac{x^2}{2} + C). Distractor A incorrectly adds C after applying the sine function. Distractors B and D use incorrect antiderivatives.

Question 8

Solve the differential equation y′=y2sin⁡(x)y' = y^2 \sin(x).

  1. y=−1cos⁡(x)+Cy = \frac{-1}{\cos(x)} + C
  2. y=−cos⁡(x)+Cy = -\cos(x) + C
  3. y=ln⁡(sec⁡(x))+Cy = \ln(\sec(x)) + C
  4. y=1cos⁡(x)+Cy = \frac{1}{\cos(x)+C} (correct answer)
Explanation: This is a separable differential equation, which means you can separate the variables and integrate both sides. When you see an equation of the form y′=f(x)g(y)y' = f(x)g(y), your strategy is to rewrite it as dydx=f(x)g(y)\frac{dy}{dx} = f(x)g(y) and then separate variables. Starting with y′=y2sin⁡(x)y' = y^2 \sin(x), rewrite as dydx=y2sin⁡(x)\frac{dy}{dx} = y^2 \sin(x). Separate variables by dividing both sides by y2y^2 and multiplying by dxdx: dyy2=sin⁡(x)dx\frac{dy}{y^2} = \sin(x) dx. Now integrate both sides: ∫y−2dy=∫sin⁡(x)dx\int y^{-2} dy = \int \sin(x) dx. The left side gives y−1−1=−1y\frac{y^{-1}}{-1} = -\frac{1}{y}, and the right side gives −cos⁡(x)-\cos(x). So you have −1y=−cos⁡(x)+C-\frac{1}{y} = -\cos(x) + C. Solving for yy: 1y=cos⁡(x)−C\frac{1}{y} = \cos(x) - C, which means y=1cos⁡(x)−Cy = \frac{1}{\cos(x) - C}. Since CC is an arbitrary constant, −C-C is also arbitrary, so we can write this as y=1cos⁡(x)+Cy = \frac{1}{\cos(x) + C}. Choice A gives y=−1cos⁡(x)+Cy = \frac{-1}{\cos(x)} + C, which has the wrong form entirely—it's not even a solution to our separated equation. Choice B, y=−cos⁡(x)+Cy = -\cos(x) + C, would give y′=sin⁡(x)y' = \sin(x), not y2sin⁡(x)y^2\sin(x). Choice C, y=ln⁡(sec⁡(x))+Cy = \ln(\sec(x)) + C, differentiates to y′=tan⁡(x)y' = \tan(x), again missing the y2y^2 term. For separable differential equations, always check your algebra when solving for the dependent variable—sign errors and reciprocal mistakes are common pitfalls.

Question 9

Find the general solution to y′=(y+1)2y' = (y+1)^2.

  1. y=tan⁡(x+C)−1y = \tan(x+C) - 1
  2. y=ln⁡((y+1)2)+Cy = \ln((y+1)^2) + C
  3. y=(C−x)−2−1y = (C-x)^{-2} - 1
  4. y=1C−x−1y = \frac{1}{C-x} - 1 (correct answer)
Explanation: When you encounter a differential equation like y′=(y+1)2y' = (y+1)^2, you're looking at a separable differential equation. The key is to separate the variables so that all terms involving yy are on one side and all terms involving xx are on the other. Starting with dydx=(y+1)2\frac{dy}{dx} = (y+1)^2, you can rewrite this as dy(y+1)2=dx\frac{dy}{(y+1)^2} = dx. Now integrate both sides. The left side requires the substitution u=y+1u = y+1, giving us ∫(y+1)−2dy=∫dx\int (y+1)^{-2} dy = \int dx, which becomes −1y+1=x+C1-\frac{1}{y+1} = x + C_1. Solving for yy, multiply both sides by −1-1: 1y+1=−x−C1\frac{1}{y+1} = -x - C_1. Let C=−C1C = -C_1 to simplify: 1y+1=C−x\frac{1}{y+1} = C - x. Therefore, y+1=1C−xy+1 = \frac{1}{C-x}, and finally y=1C−x−1y = \frac{1}{C-x} - 1. This matches answer choice D. Answer A represents the solution to y′=1+y2y' = 1 + y^2, not (y+1)2(y+1)^2. Answer B incorrectly suggests taking the natural logarithm of the right side, which isn't the proper integration technique for this separable equation. Answer C has the wrong exponent - it would result from integrating incorrectly or confusing the separation step. Remember: for separable equations of the form y′=f(y)y' = f(y), always separate variables first, then integrate. Watch out for the negative sign when integrating 1(y+1)2\frac{1}{(y+1)^2} - it's a common source of errors.

Question 10

The general solution of dydx=1y⋅xln⁡x\frac{dy}{dx} = \frac{1}{y \cdot x \ln x} is y2=f(x,C)y^2 = f(x, C). What is f(x,C)f(x, C)?

  1. 2ln⁡∣ln⁡x∣+C2\ln|\ln x| + C (correct answer)
  2. (ln⁡x)2+C(\ln x)^2 + C
  3. 2ln⁡(x2)+C2\ln(x^2) + C
  4. ln⁡(ln⁡x)+C\ln(\ln x) + C
Explanation: Separate the variables: y dy=1xln⁡xdxy\,dy = \frac{1}{x \ln x}dx. Integrate both sides. The left side is y22\frac{y^2}{2}. For the right side, use u-substitution with u=ln⁡xu = \ln x, so du=1xdxdu = \frac{1}{x}dx. The integral is ∫1udu=ln⁡∣u∣=ln⁡∣ln⁡x∣\int \frac{1}{u}du = \ln|u| = \ln|\ln x|. So, we have y22=ln⁡∣ln⁡x∣+K\frac{y^2}{2} = \ln|\ln x| + K. Multiply by 2: y2=2ln⁡∣ln⁡x∣+2Ky^2 = 2\ln|\ln x| + 2K. Let C=2KC = 2K. Thus, y2=2ln⁡∣ln⁡x∣+Cy^2 = 2\ln|\ln x| + C.

Question 11

Find the general solution to the differential equation dydx=2xy−x\frac{dy}{dx} = 2xy - x.

  1. y=Kex2−12y = K e^{x^2} - \frac{1}{2}
  2. y=12+Kex2y = \frac{1}{2} + K e^{x^2} (correct answer)
  3. y=ex2+Cy = e^{x^2} + C
  4. y=Kex2−1y = K e^{x^2-1}
Explanation: To solve the differential equation, first factor the right side to separate the variables: dydx=x(2y−1)\frac{dy}{dx} = x(2y - 1). Separate the variables by moving terms involving yy to the left and terms involving xx to the right: 12y−1dy=xdx\frac{1}{2y-1} dy = x dx. Integrate both sides: ∫12y−1dy=∫xdx\int \frac{1}{2y-1} dy = \int x dx. This yields 12ln⁡∣2y−1∣=12x2+C\frac{1}{2} \ln|2y - 1| = \frac{1}{2}x^2 + C. Multiply by 2 to simplify: ln⁡∣2y−1∣=x2+2C\ln|2y - 1| = x^2 + 2C. Let C1=2CC_1 = 2C, so ln⁡∣2y−1∣=x2+C1\ln|2y - 1| = x^2 + C_1. Exponentiate both sides: ∣2y−1∣=ex2+C1=ex2eC1|2y - 1| = e^{x^2 + C_1} = e^{x^2}e^{C_1}. Let K=±eC1K = \pm e^{C_1}, which is an arbitrary non-zero constant. We can remove the absolute value: 2y−1=Kex22y - 1 = K e^{x^2}. Finally, solve for yy: 2y=1+Kex22y = 1 + K e^{x^2}, which gives y=12+K2ex2y = \frac{1}{2} + \frac{K}{2} e^{x^2}. Since KK is an arbitrary constant, so is K/2K/2. We can rename it KK to get the final form y=12+Kex2y = \frac{1}{2} + K e^{x^2}.

Question 12

Find the general solution to the differential equation (1+x2)dy−2xydx=0(1+x^2) dy - 2xy dx = 0.

  1. y=1+x2+Cy = 1 + x^2 + C
  2. y=A(1+x2)y = A(1+x^2) (correct answer)
  3. y=Aarctan⁡(x)y = A \arctan(x)
  4. y=A1+x2y = A\sqrt{1+x^2}
Explanation: First, rearrange the equation to prepare for separation: (1+x2)dy=2xydx(1+x^2) dy = 2xy dx. Now, separate the variables by moving yy terms to the left and xx terms to the right: 1ydy=2x1+x2dx\frac{1}{y} dy = \frac{2x}{1+x^2} dx. Integrate both sides: ∫1ydy=∫2x1+x2dx\int \frac{1}{y} dy = \int \frac{2x}{1+x^2} dx. The left integral is ln⁡∣y∣\ln|y|. For the right integral, use u-substitution with u=1+x2u = 1+x^2, so du=2xdxdu = 2x dx. The integral becomes ∫1udu=ln⁡∣u∣=ln⁡(1+x2)\int \frac{1}{u} du = \ln|u| = \ln(1+x^2) (absolute value is not needed since 1+x2>01+x^2 > 0). So, ln⁡∣y∣=ln⁡(1+x2)+C\ln|y| = \ln(1+x^2) + C. Exponentiate both sides: ∣y∣=eln⁡(1+x2)+C=eln⁡(1+x2)eC=(1+x2)eC|y| = e^{\ln(1+x^2)+C} = e^{\ln(1+x^2)}e^C = (1+x^2)e^C. Let A=±eCA = \pm e^C. The solution is y=A(1+x2)y = A(1+x^2). Distractor A represents the common error of treating ea+be^{a+b} as ea+ebe^a+e^b. Distractor C results from incorrectly integrating 2x1+x2\frac{2x}{1+x^2} as if it were 11+x2\frac{1}{1+x^2}.

Question 13

Given dydx=ysin⁡(x)\frac{dy}{dx} = \sqrt{y} \sin(x) for y>0y > 0, find the general solution for yy.

  1. y=(C+cos⁡(x))2y = (C + \cos(x))^2
  2. y=14(C−cos⁡(x))2y = \frac{1}{4}(C - \cos(x))^2
  3. y=(K−12cos⁡(x))2y = (K - \frac{1}{2}\cos(x))^2 (correct answer)
  4. y=(C−2cos⁡(x))2y = (C - 2\cos(x))^2
Explanation: Separate the variables: 1ydy=sin⁡(x)dx\frac{1}{\sqrt{y}} dy = \sin(x) dx, or y−1/2dy=sin⁡(x)dxy^{-1/2} dy = \sin(x) dx. Integrate both sides: ∫y−1/2dy=∫sin⁡(x)dx\int y^{-1/2} dy = \int \sin(x) dx. Using the power rule for the left side and the standard integral for the right side, we get: 2y1/2=−cos⁡(x)+C12y^{1/2} = -\cos(x) + C_1. Now, solve for yy. First, divide by 2: y1/2=−12cos⁡(x)+C12y^{1/2} = -\frac{1}{2}\cos(x) + \frac{C_1}{2}. Let K=C12K = \frac{C_1}{2} be the new arbitrary constant. The equation is y=K−12cos⁡(x)\sqrt{y} = K - \frac{1}{2}\cos(x). Finally, square both sides to get the explicit solution: y=(K−12cos⁡(x))2y = (K - \frac{1}{2}\cos(x))^2. Distractor A has a sign error on the integral of sin⁡(x)\sin(x). Distractor B incorrectly handles the constant 2 from the integration. Distractor D incorrectly multiplies by 2 instead of dividing.

Question 14

Which equation implicitly defines the general solution to cos⁡(y)dydx=3x2\cos(y) \frac{dy}{dx} = 3x^2?

  1. tan⁡(y)=x3+C\tan(y) = x^3 + C
  2. −sin⁡(y)=x3+C-\sin(y) = x^3 + C
  3. cos⁡(y)=x3+C\cos(y) = x^3 + C
  4. sin⁡(y)=x3+C\sin(y) = x^3 + C (correct answer)
Explanation: This is a separable differential equation problem where you need to separate variables and integrate both sides to find the general solution. Starting with cos⁡(y)dydx=3x2\cos(y) \frac{dy}{dx} = 3x^2, you can rearrange to separate the variables: cos⁡(y) dy=3x2 dx\cos(y) \, dy = 3x^2 \, dx. Now integrate both sides. The left side gives you ∫cos⁡(y) dy=sin⁡(y)\int \cos(y) \, dy = \sin(y), and the right side gives you ∫3x2 dx=x3\int 3x^2 \, dx = x^3. Adding the constant of integration, you get sin⁡(y)=x3+C\sin(y) = x^3 + C, which matches answer choice D. Let's examine why the other options are incorrect. Choice A suggests tan⁡(y)=x3+C\tan(y) = x^3 + C, but if you differentiate this implicitly, you get sec⁡2(y)dydx=3x2\sec^2(y) \frac{dy}{dx} = 3x^2, not the original equation with cos⁡(y)\cos(y). Choice B gives −sin⁡(y)=x3+C-\sin(y) = x^3 + C, which would result from incorrectly integrating cos⁡(y)\cos(y) as −sin⁡(y)-\sin(y) instead of +sin⁡(y)+\sin(y). Choice C suggests cos⁡(y)=x3+C\cos(y) = x^3 + C, but this would come from mistakenly thinking that integrating cos⁡(y)\cos(y) gives cos⁡(y)\cos(y) rather than sin⁡(y)\sin(y). Study tip: When solving separable differential equations, always remember that ∫cos⁡(y) dy=sin⁡(y)+C\int \cos(y) \, dy = \sin(y) + C and ∫sin⁡(y) dy=−cos⁡(y)+C\int \sin(y) \, dy = -\cos(y) + C. After finding your solution, you can verify it by differentiating implicitly to check if you recover the original differential equation.

Question 15

The general solution for y′=kyy' = ky is y=Cekxy = Ce^{kx}. This arises from integrating 1ydy=k dx\frac{1}{y}dy = k\,dx to get ln⁡∣y∣=kx+C1\ln|y| = kx+C_1. Which statement correctly describes the relationship between the constant of integration C1C_1 and the coefficient CC?

  1. CC is any real number, derived from C=±eC1C = \pm e^{C_1}. (correct answer)
  2. CC is any positive number, derived from C=eC1C = e^{C_1}.
  3. CC is the same as C1C_1, as constants can be relabeled.
  4. CC is any real number, derived from C=C1+ln⁡∣y∣C = C_1 + \ln|y|.
Explanation: Starting from ln⁡∣y∣=kx+C1\ln|y| = kx+C_1, we exponentiate both sides to get ∣y∣=ekx+C1=eC1ekx|y| = e^{kx+C_1} = e^{C_1}e^{kx}. The absolute value means y=±eC1ekxy = \pm e^{C_1}e^{kx}. Since eC1e^{C_1} must be positive, ±eC1\pm e^{C_1} can be any non-zero real number. By also considering the trivial solution y=0y=0, the constant CC in the general solution y=Cekxy = Ce^{kx} can be any real number.

Question 16

Find the general solution, in implicit form, for the differential equation (1+x2)dydx+y2=0(1+x^2) \frac{dy}{dx} + y^2 = 0.

  1. 1y+arctan⁡(x)=C\frac{1}{y} + \arctan(x) = C (correct answer)
  2. −1y+ln⁡(1+x2)=C-\frac{1}{y} + \ln(1+x^2) = C
  3. ln⁡(y2)+ln⁡(1+x2)=C\ln(y^2) + \ln(1+x^2) = C
  4. arctan⁡(y)−1x=C\arctan(y) - \frac{1}{x} = C
Explanation: First, rearrange the equation to separate variables: (1+x2)dydx=−y2(1+x^2) \frac{dy}{dx} = -y^2. Then, 1y2dy=−11+x2dx\frac{1}{y^2}dy = -\frac{1}{1+x^2}dx. Integrate both sides: ∫y−2dy=−∫11+x2dx\int y^{-2}dy = -\int \frac{1}{1+x^2}dx. This yields −y−1=−arctan⁡(x)+K-y^{-1} = -\arctan(x) + K. Multiplying by -1 gives 1y=arctan⁡(x)−K\frac{1}{y} = \arctan(x) - K. Let C=−KC = -K. The solution is 1y=arctan⁡(x)+C\frac{1}{y} = \arctan(x) + C, which can be rewritten as 1y−arctan⁡(x)=C\frac{1}{y} - \arctan(x) = C. Choice A is equivalent by renaming the constant C.

Question 17

Find the general solution for yy given the differential equation dydx=y2+1x2\frac{dy}{dx} = \frac{y^2+1}{x^2}.

  1. y=tan⁡(C−1x)y = \tan(C - \frac{1}{x}) (correct answer)
  2. y=tan⁡(C)−tan⁡(1x)y = \tan(C) - \tan(\frac{1}{x})
  3. y=arctan⁡(C−1x)y = \arctan(C - \frac{1}{x})
  4. y=tan⁡(ln⁡∣x2∣+C)y = \tan(\ln|x^2| + C)
Explanation: First, separate the variables: 1y2+1dy=1x2dx\frac{1}{y^2+1} dy = \frac{1}{x^2} dx. Next, integrate both sides: ∫1y2+1dy=∫x−2dx\int \frac{1}{y^2+1} dy = \int x^{-2} dx. The integral of the left side is arctan⁡(y)\arctan(y). The integral of the right side is −x−1+C-x^{-1} + C, or C−1xC - \frac{1}{x}. So, we have the equation arctan⁡(y)=C−1x\arctan(y) = C - \frac{1}{x}. To solve for yy, we take the tangent of both sides: y=tan⁡(C−1x)y = \tan(C - \frac{1}{x}). Distractor B incorrectly distributes the tangent function. Distractor C forgets to apply the tangent function to solve for yy. Distractor D uses an incorrect antiderivative for 1/x21/x^2.

Question 18

What is the general solution to the differential equation dydx=ex+y\frac{dy}{dx} = e^{x+y}?

  1. y=−ln⁡(C−ex)y = -\ln(C - e^x) (correct answer)
  2. y=−ln⁡(ex+C)y = -\ln(e^x + C)
  3. y=x+Cy = x + C
  4. y=ex+ey+Cy = e^x + e^y + C
Explanation: First, use the property of exponents to rewrite the equation: dydx=exey\frac{dy}{dx} = e^x e^y. Separate the variables by dividing by eye^y: 1eydy=exdx\frac{1}{e^y} dy = e^x dx, which is e−ydy=exdxe^{-y} dy = e^x dx. Integrate both sides: ∫e−ydy=∫exdx\int e^{-y} dy = \int e^x dx. This gives −e−y=ex+C1-e^{-y} = e^x + C_1. To solve for yy, first multiply by -1: e−y=−ex−C1e^{-y} = -e^x - C_1. Let C=−C1C = -C_1. Then e−y=C−exe^{-y} = C - e^x. Take the natural logarithm of both sides: −y=ln⁡(C−ex)-y = \ln(C - e^x). Finally, multiply by -1: y=−ln⁡(C−ex)y = -\ln(C - e^x). Distractor B results from a sign error when integrating e−ye^{-y}. Distractors C and D result from fundamental misunderstandings of how to solve this type of differential equation.

Question 19

The general solution to the differential equation y′=yxln⁡xy' = \frac{y}{x \ln x} can be written in the form y=f(x)y = f(x). What is f(x)f(x)?

  1. f(x)=ln⁡(x)+Cf(x) = \ln(x) + C
  2. f(x)=Aln⁡(x)f(x) = A \ln(x) (correct answer)
  3. f(x)=A(ln⁡x)2f(x) = A(\ln x)^2
  4. f(x)=ex+Cf(x) = e^{x} + C
Explanation: First, write the equation as dydx=yxln⁡x\frac{dy}{dx} = \frac{y}{x \ln x} and separate the variables: 1ydy=1xln⁡xdx\frac{1}{y} dy = \frac{1}{x \ln x} dx. Next, integrate both sides: ∫1ydy=∫1xln⁡xdx\int \frac{1}{y} dy = \int \frac{1}{x \ln x} dx. The left integral is ln⁡∣y∣\ln|y|. For the right integral, use u-substitution with u=ln⁡xu = \ln x, so du=1xdxdu = \frac{1}{x} dx. The integral becomes ∫1udu=ln⁡∣u∣=ln⁡∣ln⁡x∣\int \frac{1}{u} du = \ln|u| = \ln|\ln x|. So, we have ln⁡∣y∣=ln⁡∣ln⁡x∣+C\ln|y| = \ln|\ln x| + C. To solve for yy, exponentiate both sides: ∣y∣=eln⁡∣ln⁡x∣+C=eln⁡∣ln⁡x∣⋅eC=∣ln⁡x∣⋅eC|y| = e^{\ln|\ln x| + C} = e^{\ln|\ln x|} \cdot e^C = |\ln x| \cdot e^C. Let A=±eCA = \pm e^C. Then y=Aln⁡(x)y = A \ln(x).

Question 20

Find the general solution, in implicit form, to the differential equation xydydx=y+1xy \frac{dy}{dx} = y+1.

  1. y−ln⁡∣y+1∣=ln⁡∣x∣+Cy - \ln|y+1| = \ln|x| + C (correct answer)
  2. y+ln⁡∣y+1∣=ln⁡∣x∣+Cy + \ln|y+1| = \ln|x| + C
  3. y22=ln⁡∣x∣+ln⁡∣y+1∣+C\frac{y^2}{2} = \ln|x| + \ln|y+1| + C
  4. y−ln⁡∣y+1∣=−1x2+Cy - \ln|y+1| = -\frac{1}{x^2} + C
Explanation: To separate the variables, move all yy terms to the left and all xx terms to the right: yy+1dy=1xdx\frac{y}{y+1} dy = \frac{1}{x} dx. To integrate the left side, ∫yy+1dy\int \frac{y}{y+1} dy, we can rewrite the integrand as y+1−1y+1=1−1y+1\frac{y+1-1}{y+1} = 1 - \frac{1}{y+1}. Now integrate both sides: ∫(1−1y+1)dy=∫1xdx\int (1 - \frac{1}{y+1}) dy = \int \frac{1}{x} dx. This yields y−ln⁡∣y+1∣=ln⁡∣x∣+Cy - \ln|y+1| = \ln|x| + C. This is the implicit general solution. Distractor B has a sign error in the integration of the yy term. Distractor C results from an incorrect attempt to integrate yy+1\frac{y}{y+1}. Distractor D uses the wrong antiderivative for 1x\frac{1}{x}.