Differential Equations Quiz: Characteristic Equation
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Characteristic EquationQuestion 1 of 20

The function y1(t)=e2tsin⁡(5t)y_1(t) = e^{2t}\sin(5t) is a solution to a certain second-order, linear, homogeneous differential equation with real constant coefficients. Which of the following functions must also be a solution to this equation?

y(t)=e2tcos⁡(5t)y(t) = e^{2t}\cos(5t)
y(t)=e−2tcos⁡(5t)y(t) = e^{-2t}\cos(5t)
y(t)=te2tsin⁡(5t)y(t) = t e^{2t}\sin(5t)
y(t)=e5tsin⁡(2t)y(t) = e^{5t}\sin(2t)
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Differential Equations Quiz

Differential Equations Quiz: Characteristic Equation

Practice Characteristic Equation in Differential Equations 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 Characteristic Equation, giving you a quick way to practice the rules, question types, and explanations that matter most for Differential Equations.

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

The function y1(t)=e2tsin⁡(5t)y_1(t) = e^{2t}\sin(5t) is a solution to a certain second-order, linear, homogeneous differential equation with real constant coefficients. Which of the following functions must also be a solution to this equation?

  1. y(t)=e2tcos⁡(5t)y(t) = e^{2t}\cos(5t) (correct answer)
  2. y(t)=e−2tcos⁡(5t)y(t) = e^{-2t}\cos(5t)
  3. y(t)=te2tsin⁡(5t)y(t) = t e^{2t}\sin(5t)
  4. y(t)=e5tsin⁡(2t)y(t) = e^{5t}\sin(2t)
Explanation: For a differential equation with real coefficients, if a complex-valued function is a solution, its conjugate must also be a solution. The given solution e2tsin⁡(5t)e^{2t}\sin(5t) arises from characteristic roots r=2±5ir = 2 \pm 5i. The general solution is of the form y(t)=e2t(c1cos⁡(5t)+c2sin⁡(5t))y(t) = e^{2t}(c_1\cos(5t) + c_2\sin(5t)). Since y1(t)=e2tsin⁡(5t)y_1(t) = e^{2t}\sin(5t) is a solution (corresponding to c1=0,c2=1c_1=0, c_2=1), the other linearly independent basis function, y2(t)=e2tcos⁡(5t)y_2(t) = e^{2t}\cos(5t), must also be a solution (corresponding to c1=1,c2=0c_1=1, c_2=0).

Question 2

Consider the equation 2y′′+by′+8y=02y'' + by' + 8y = 0, where b≥0b \ge 0. The behavior of the solutions transitions from oscillatory to non-oscillatory as bb increases. At what value of bb does this transition occur?

  1. b=4b=4
  2. b=8b=8 (correct answer)
  3. b=16b=16
  4. b=32b=32
Explanation: The transition from oscillatory to non-oscillatory behavior occurs when the roots of the characteristic equation change from complex conjugate to real. This happens at the point of critical damping, where the discriminant is zero. The characteristic equation is 2r2+br+8=02r^2 + br + 8 = 0. The discriminant is D=b2−4ac=b2−4(2)(8)=b2−64D = b^2 - 4ac = b^2 - 4(2)(8) = b^2 - 64. Setting the discriminant to zero gives b2−64=0b^2 - 64 = 0. Since the problem states b≥0b \ge 0, we take the positive root, b=8b=8. For 0≤b<80 \le b < 8, the roots are complex (oscillatory), and for b>8b > 8, the roots are distinct real (non-oscillatory). The transition occurs at b=8b=8.

Question 3

Two linearly independent solutions to y′′+py′+qy=0y'' + py' + qy = 0 are y1(t)=eatcos⁡(bt)y_1(t) = e^{at}\cos(bt) and y2(t)=eatsin⁡(bt)y_2(t) = e^{at}\sin(bt), where aa and bb are non-zero real numbers. What is the value of qq in terms of aa and bb?

  1. q=−2aq = -2a
  2. q=a2−b2q = a^2 - b^2
  3. q=a2+b2q = a^2 + b^2 (correct answer)
  4. q=b2−a2q = b^2 - a^2
Explanation: When you encounter a second-order linear homogeneous differential equation with complex exponential solutions like eatcos⁡(bt)e^{at}\cos(bt) and eatsin⁡(bt)e^{at}\sin(bt), you're dealing with a characteristic equation that has complex roots. This form tells you immediately that the characteristic roots are r=a±bir = a \pm bi. For the differential equation y′′+py′+qy=0y'' + py' + qy = 0, the characteristic equation is r2+pr+q=0r^2 + pr + q = 0. Since we know the roots are r1=a+bir_1 = a + bi and r2=a−bir_2 = a - bi, we can use Vieta's formulas to find the coefficients. The sum of roots gives us r1+r2=(a+bi)+(a−bi)=2a=−pr_1 + r_2 = (a + bi) + (a - bi) = 2a = -p, so p=−2ap = -2a. The product of roots gives us r1⋅r2=(a+bi)(a−bi)=a2−(bi)2=a2−b2i2=a2+b2=qr_1 \cdot r_2 = (a + bi)(a - bi) = a^2 - (bi)^2 = a^2 - b^2i^2 = a^2 + b^2 = q. Therefore, q=a2+b2q = a^2 + b^2, which is answer C. Looking at the wrong answers: A) q=−2aq = -2a confuses the coefficient qq with pp; this is actually what −p-p equals. B) q=a2−b2q = a^2 - b^2 makes the common error of forgetting that i2=−1i^2 = -1 when expanding (a+bi)(a−bi)(a + bi)(a - bi). D) q=b2−a2q = b^2 - a^2 reverses the correct relationship and also misses the sign change from i2i^2. Remember: when you see exponential-trigonometric solutions, the characteristic roots are complex conjugates a±bia \pm bi, and their product always gives a2+b2a^2 + b^2 for the constant term.

Question 4

The motion of a mechanical system is described by 3y′′+15y=−12y′3y'' + 15y = -12y'. Which of the following is the general solution y(t)y(t) for the system's displacement?

  1. y(t)=e−2t(c1cos⁡(t)+c2sin⁡(t))y(t) = e^{-2t}(c_1 \cos(t) + c_2 \sin(t)) (correct answer)
  2. y(t)=e2t(c1cos⁡(t)+c2sin⁡(t))y(t) = e^{2t}(c_1 \cos(t) + c_2 \sin(t))
  3. y(t)=c1cos⁡(5t)+c2sin⁡(5t)y(t) = c_1 \cos(\sqrt{5}t) + c_2 \sin(\sqrt{5}t)
  4. y(t)=e−t(c1cos⁡(2t)+c2sin⁡(2t))y(t) = e^{-t}(c_1 \cos(2t) + c_2 \sin(2t))
Explanation: First, rewrite the differential equation in the standard form ay′′+by′+cy=0ay'' + by' + cy = 0. Adding 12y′12y' to both sides gives 3y′′+12y′+15y=03y'' + 12y' + 15y = 0. Divide the entire equation by 3 to simplify: y′′+4y′+5y=0y'' + 4y' + 5y = 0. The characteristic equation is r2+4r+5=0r^2 + 4r + 5 = 0. Using the quadratic formula, the roots are r=−4±42−4(1)(5)2(1)=−4±16−202=−4±−42=−4±2i2=−2±ir = \frac{-4 \pm \sqrt{4^2 - 4(1)(5)}}{2(1)} = \frac{-4 \pm \sqrt{16 - 20}}{2} = \frac{-4 \pm \sqrt{-4}}{2} = \frac{-4 \pm 2i}{2} = -2 \pm i. The roots are a complex conjugate pair α±iβ\alpha \pm i\beta with α=−2\alpha=-2 and β=1\beta=1. The general solution is y(t)=eαt(c1cos⁡(βt)+c2sin⁡(βt))y(t) = e^{\alpha t}(c_1 \cos(\beta t) + c_2 \sin(\beta t)), which is y(t)=e−2t(c1cos⁡(t)+c2sin⁡(t))y(t) = e^{-2t}(c_1 \cos(t) + c_2 \sin(t)).

Question 5

Let y1(t)y_1(t) be the solution to y′′+9y=0y'' + 9y = 0 with y(0)=1,y′(0)=0y(0)=1, y'(0)=0, and let y2(t)y_2(t) be the solution to y′′+6y′+9y=0y'' + 6y' + 9y = 0 with y(0)=1,y′(0)=0y(0)=1, y'(0)=0. Which statement correctly compares the forms of their characteristic roots?

  1. The roots for y1y_1 are a repeated real number, while the roots for y2y_2 are distinct real numbers.
  2. The roots for y1y_1 are pure imaginary, while the roots for y2y_2 are distinct real numbers.
  3. The roots for y1y_1 are a repeated real number, while the roots for y2y_2 are complex conjugates.
  4. The roots for y1y_1 are pure imaginary, while the roots for y2y_2 are a repeated real number. (correct answer)
Explanation: When you encounter second-order linear homogeneous differential equations, the key is finding the characteristic roots, which determine the form of your solution. You solve this by substituting y=erty = e^{rt} and finding the roots of the resulting characteristic equation. For y1(t)y_1(t): The equation y′′+9y=0y'' + 9y = 0 gives the characteristic equation r2+9=0r^2 + 9 = 0, so r2=−9r^2 = -9. This yields r=±3ir = \pm 3i, which are pure imaginary numbers (no real part, only imaginary components). For y2(t)y_2(t): The equation y′′+6y′+9y=0y'' + 6y' + 9y = 0 gives r2+6r+9=0r^2 + 6r + 9 = 0. This factors as (r+3)2=0(r + 3)^2 = 0, yielding the repeated real root r=−3r = -3. Looking at the answer choices: Choice A incorrectly reverses which equation has repeated roots and gets the type wrong for y1y_1. Choice B wrongly claims y2y_2 has distinct real roots when it actually has repeated roots. Choice C incorrectly states y1y_1 has repeated real roots and y2y_2 has complex conjugates. Choice D correctly identifies that y1y_1 has pure imaginary roots (±3i\pm 3i) while y2y_2 has a repeated real root (−3-3). Study tip: Remember the pattern for characteristic equations: if the discriminant is negative, you get complex/imaginary roots; if it's zero, you get repeated real roots; if it's positive, you get distinct real roots. Also, pure imaginary means the real part is zero.

Question 6

The differential equation y′′−10y′+cy=0y'' - 10y' + c y = 0 has a general solution of the form y(t)=c1er1t+c2er2ty(t) = c_1 e^{r_1 t} + c_2 e^{r_2 t} where r1r_1 and r2r_2 are distinct real numbers. If one root of the characteristic equation is r1=4r_1 = 4, what is the general solution?

  1. y(t)=c1e4t+c2te4ty(t) = c_1 e^{4t} + c_2 t e^{4t}
  2. y(t)=c1e4t+c2e−14ty(t) = c_1 e^{4t} + c_2 e^{-14t}
  3. y(t)=c1e4t+c2e6ty(t) = c_1 e^{4t} + c_2 e^{6t} (correct answer)
  4. y(t)=c1e4t+c2e5ty(t) = c_1 e^{4t} + c_2 e^{5t}
Explanation: When you encounter a second-order linear homogeneous differential equation like y′′−10y′+cy=0y'' - 10y' + cy = 0, you solve it by finding the characteristic equation and its roots. The characteristic equation is formed by substituting y=erty = e^{rt}, giving you r2−10r+c=0r^2 - 10r + c = 0. Since you're told that one root is r1=4r_1 = 4 and the roots are distinct real numbers, you can find the second root using the relationship between roots and coefficients. For a quadratic r2−10r+c=0r^2 - 10r + c = 0, the sum of roots equals 10 (the negative of the coefficient of rr). Therefore: r1+r2=10r_1 + r_2 = 10, so 4+r2=104 + r_2 = 10, which gives r2=6r_2 = 6. The general solution is y(t)=c1e4t+c2e6ty(t) = c_1 e^{4t} + c_2 e^{6t}, which is answer C. Let's examine why the other options are incorrect: A) y(t)=c1e4t+c2te4ty(t) = c_1 e^{4t} + c_2 t e^{4t} represents the case of repeated roots, not distinct roots as specified in the problem. B) y(t)=c1e4t+c2e−14ty(t) = c_1 e^{4t} + c_2 e^{-14t} would require roots that sum to 4+(−14)=−104 + (-14) = -10, but our coefficient pattern requires a sum of +10. D) y(t)=c1e4t+c2e5ty(t) = c_1 e^{4t} + c_2 e^{5t} gives roots summing to 9, not 10 as required by the characteristic equation. Study tip: Always use Vieta's formulas when you know one root of a quadratic characteristic equation. For r2+br+c=0r^2 + br + c = 0, the sum of roots equals −b-b and the product equals cc. This saves time compared to solving the full quadratic.

Question 7

Consider the second-order linear homogeneous ODE ay′′+by′+cy=0ay'' + by' + cy = 0 where aa, bb, and cc are real constants with a≠0a \neq 0. If the characteristic equation has roots r1=2+3ir_1 = 2 + 3i and r2=2−3ir_2 = 2 - 3i, which of the following must be true about the coefficients?

  1. b=−4ab = -4a and c=13ac = 13a (correct answer)
  2. b=4ab = 4a and c=−13ac = -13a
  3. b=−4ab = -4a and c=−13ac = -13a
  4. b=4ab = 4a and c=13ac = 13a
Explanation: For a quadratic equation ar2+br+c=0ar^2 + br + c = 0 with roots r1=2+3ir_1 = 2 + 3i and r2=2−3ir_2 = 2 - 3i, we use Vieta's formulas. The sum of roots is r1+r2=(2+3i)+(2−3i)=4=−bar_1 + r_2 = (2 + 3i) + (2 - 3i) = 4 = -\frac{b}{a}, so b=−4ab = -4a. The product of roots is r1⋅r2=(2+3i)(2−3i)=4−9i2=4+9=13=car_1 \cdot r_2 = (2 + 3i)(2 - 3i) = 4 - 9i^2 = 4 + 9 = 13 = \frac{c}{a}, so c=13ac = 13a. Choice B has the wrong sign for bb. Choice C has the wrong sign for cc. Choice D has wrong signs for both coefficients.

Question 8

For the differential equation 4y′′+4y′+ky=04y'' + 4y' + ky = 0, determine the value of kk such that the characteristic equation has roots with real part equal to −12-\frac{1}{2} and the general solution contains oscillatory terms.

  1. k=0k = 0
  2. k=1k = 1
  3. k=2k = 2 (correct answer)
  4. k=4k = 4
Explanation: The characteristic equation is 4r2+4r+k=04r^2 + 4r + k = 0, or r2+r+k4=0r^2 + r + \frac{k}{4} = 0. Using the quadratic formula: r=−1±1−k2r = \frac{-1 \pm \sqrt{1 - k}}{2}. For oscillatory terms, we need complex roots, so 1−k<01 - k < 0, meaning k>1k > 1. For real part −12-\frac{1}{2}, we need r=−12±bir = -\frac{1}{2} \pm bi where b≠0b \neq 0. From the formula, the real part is −12-\frac{1}{2} (which matches) and b=k−12b = \frac{\sqrt{k-1}}{2}. For oscillatory behavior, we need k>1k > 1. Among the choices greater than 1, we check: if k=2k = 2, then r=−12±i2r = -\frac{1}{2} \pm \frac{i}{2}, giving oscillatory solutions. Choice A gives repeated roots, choice B gives the boundary case, and choice D gives a different real part.

Question 9

Consider two second-order linear ODEs with characteristic equations r2+ar+b=0r^2 + ar + b = 0 and r2+cr+d=0r^2 + cr + d = 0. If the first equation has roots 2±3i2 \pm 3i and the second has roots −2±3i-2 \pm 3i, what is the relationship between the coefficients?

  1. a−c=4a - c = 4 and b−d=0b - d = 0
  2. a=ca = c and b+d=0b + d = 0
  3. a+c=0a + c = 0 and b+d=26b + d = 26
  4. a+c=0a + c = 0 and b=db = d (correct answer)
Explanation: When you encounter characteristic equations for second-order linear ODEs, remember that the coefficients are directly related to the roots through Vieta's formulas. For a quadratic r2+pr+q=0r^2 + pr + q = 0 with roots r1r_1 and r2r_2, we have r1+r2=−pr_1 + r_2 = -p and r1⋅r2=qr_1 \cdot r_2 = q. For the first equation with roots 2±3i2 \pm 3i, the sum is (2+3i)+(2−3i)=4(2 + 3i) + (2 - 3i) = 4, so a=−4a = -4. The product is (2+3i)(2−3i)=4−9i2=4+9=13(2 + 3i)(2 - 3i) = 4 - 9i^2 = 4 + 9 = 13, so b=13b = 13. For the second equation with roots −2±3i-2 \pm 3i, the sum is (−2+3i)+(−2−3i)=−4(-2 + 3i) + (-2 - 3i) = -4, so c=4c = 4. The product is (−2+3i)(−2−3i)=4−9i2=13(-2 + 3i)(-2 - 3i) = 4 - 9i^2 = 13, so d=13d = 13. Therefore, a+c=−4+4=0a + c = -4 + 4 = 0 and b=d=13b = d = 13. Choice A incorrectly calculates a−c=−4−4=−8a - c = -4 - 4 = -8, not 4, and while b−d=0b - d = 0 is true, the first condition fails. Choice B claims a=ca = c, but −4≠4-4 \neq 4, and b+d=26b + d = 26, not 0. Choice C correctly identifies a+c=0a + c = 0 but incorrectly states b+d=26b + d = 26 instead of recognizing that b=db = d. Study tip: Always use Vieta's formulas systematically: sum of roots equals the negative of the linear coefficient, and product of roots equals the constant term. This eliminates calculation errors and reveals relationships between different characteristic equations.

Question 10

The characteristic equation r2−6r+k=0r^2 - 6r + k = 0 corresponds to a second-order linear ODE. For which value of kk will the general solution contain terms of the form e3x(Acos⁡(2x)+Bsin⁡(2x))e^{3x}(A\cos(2x) + B\sin(2x))?

  1. k=5k = 5
  2. k=9k = 9
  3. k=13k = 13 (correct answer)
  4. k=17k = 17
Explanation: The given solution form e3x(Acos⁡(2x)+Bsin⁡(2x))e^{3x}(A\cos(2x) + B\sin(2x)) indicates complex conjugate roots of the form 3±2i3 \pm 2i. Using the quadratic formula on r2−6r+k=0r^2 - 6r + k = 0: r=6±36−4k2=3±36−4k2r = \frac{6 \pm \sqrt{36 - 4k}}{2} = 3 \pm \frac{\sqrt{36 - 4k}}{2}. For roots 3±2i3 \pm 2i, we need 36−4k2=2i\frac{\sqrt{36 - 4k}}{2} = 2i, which means 36−4k=−1636 - 4k = -16, so k=13k = 13. Choice A gives real distinct roots, choice B gives repeated real roots, and choice D gives complex roots with wrong imaginary part.

Question 11

The general solution of a second-order linear homogeneous ODE is given by y(x)=c1e−2x+c2xe−2xy(x) = c_1 e^{-2x} + c_2 xe^{-2x}. If this same ODE is written in the standard form y′′+py′+qy=0y'' + py' + qy = 0, what is the value of the discriminant of its characteristic equation?

  1. The discriminant equals p2−4q=16p^2 - 4q = 16
  2. The discriminant equals p2−4q=0p^2 - 4q = 0 (correct answer)
  3. The discriminant equals p2−4q=−4p^2 - 4q = -4
  4. The discriminant equals p2−4q=4p^2 - 4q = 4
Explanation: The solution y(x)=c1e−2x+c2xe−2xy(x) = c_1 e^{-2x} + c_2 xe^{-2x} indicates a repeated root r=−2r = -2 in the characteristic equation r2+pr+q=0r^2 + pr + q = 0. For a repeated root, the characteristic equation must be (r+2)2=r2+4r+4=0(r + 2)^2 = r^2 + 4r + 4 = 0, so p=4p = 4 and q=4q = 4. The discriminant is p2−4q=16−16=0p^2 - 4q = 16 - 16 = 0, which confirms the repeated root case. Choices A and D suggest distinct real roots, while choice C suggests complex roots.

Question 12

For the differential equation y′′−4y′+ky=0y'' - 4y' + ky = 0, the characteristic equation has two distinct real roots when k<4k < 4, repeated real roots when k=4k = 4, and complex conjugate roots when k>4k > 4. If a particular solution is known to be yp(x)=x2e2xy_p(x) = x^2 e^{2x}, what can be concluded about the value of kk?

  1. k=4k = 4 and the characteristic equation has a double root at r=2r = 2 (correct answer)
  2. k<4k < 4 and the characteristic equation has distinct roots r=0,4r = 0, 4
  3. k>4k > 4 and the characteristic equation has complex roots 2±ik−42 \pm i\sqrt{k-4}
  4. k=0k = 0 and the characteristic equation has distinct roots r=0,4r = 0, 4
Explanation: The solution form yp(x)=x2e2xy_p(x) = x^2 e^{2x} suggests that r=2r = 2 is a repeated root of multiplicity at least 2 (since we have x2x^2 as a factor). For r2−4r+k=0r^2 - 4r + k = 0, if r=2r = 2 is a double root, then (r−2)2=r2−4r+4=0(r-2)^2 = r^2 - 4r + 4 = 0, so k=4k = 4. This matches the condition for repeated roots. Choice B incorrectly assumes distinct roots, choice C incorrectly assumes complex roots, and choice D gives the wrong value of kk.

Question 13

The characteristic equation for ay′′+by′+cy=0ay''+by'+cy=0 is ar2+br+c=0ar^2+br+c=0. If a>0a>0, c>0c>0, and b=4acb = \sqrt{4ac}, which of the following best describes the non-trivial solutions y(t)y(t)?

  1. Solutions are oscillatory with constant amplitude.
  2. Solutions are non-oscillatory and decay to zero as t→∞t \to \infty. (correct answer)
  3. Solutions are non-oscillatory and grow unboundedly as t→∞t \to \infty.
  4. Solutions are oscillatory with exponentially decaying amplitude.
Explanation: The condition b=4acb = \sqrt{4ac} is equivalent to b2=4acb^2 = 4ac, which means the discriminant b2−4acb^2 - 4ac is zero. This indicates a repeated real root for the characteristic equation. The root is given by r=−b2ar = \frac{-b}{2a}. Since a>0a>0 and b=4acb=\sqrt{4ac} must be positive (as a,c>0a,c>0), the root r=−b2ar = \frac{-b}{2a} is negative. The general solution is y(t)=(c1+c2t)erty(t) = (c_1 + c_2 t)e^{rt}. Since r<0r<0, the exponential term erte^{rt} goes to zero as t→∞t \to \infty. Although the term c2tc_2t grows, the exponential decay is dominant, so lim⁡t→∞y(t)=0\lim_{t \to \infty} y(t) = 0. The solution is non-oscillatory because the roots are real.

Question 14

The general solution to a constant-coefficient, homogeneous second-order linear differential equation is given by y(t)=c1e−3t+c2te−3ty(t) = c_1 e^{-3t} + c_2 t e^{-3t}. What is the differential equation?

  1. y′′−6y′+9y=0y'' - 6y' + 9y = 0
  2. y′′+6y′+9y=0y'' + 6y' + 9y = 0 (correct answer)
  3. y′′+2y′−3y=0y'' + 2y' - 3y = 0
  4. y′′+3y′=0y'' + 3y' = 0
Explanation: The form of the general solution y(t)=c1ert+c2terty(t) = c_1 e^{rt} + c_2 t e^{rt} indicates a repeated real root rr for the characteristic equation. Here, the solution is y(t)=c1e−3t+c2te−3ty(t) = c_1 e^{-3t} + c_2 t e^{-3t}, so the repeated root is r=−3r = -3. The characteristic equation must be (r−(−3))2=0(r - (-3))^2 = 0, which expands to (r+3)2=0(r+3)^2 = 0, or r2+6r+9=0r^2 + 6r + 9 = 0. This characteristic equation corresponds to the differential equation y′′+6y′+9y=0y'' + 6y' + 9y = 0.

Question 15

Consider the differential equation y′′+βy′+9y=0y'' + \beta y' + 9y = 0. For which values of the real parameter β\beta will all non-trivial solutions exhibit oscillatory behavior?

  1. β>6\beta > 6 or β<−6\beta < -6
  2. β=6\beta = 6 or β=−6\beta = -6
  3. −6<β<6-6 < \beta < 6 (correct answer)
  4. 0<β<60 < \beta < 6
Explanation: Oscillatory behavior in the solutions of a second-order linear homogeneous ODE corresponds to complex conjugate roots of its characteristic equation. The characteristic equation is r2+βr+9=0r^2 + \beta r + 9 = 0. The roots are complex when the discriminant is negative. The discriminant is D=b2−4ac=β2−4(1)(9)=β2−36D = b^2 - 4ac = \beta^2 - 4(1)(9) = \beta^2 - 36. For complex roots, we need D<0D < 0, which means β2−36<0\beta^2 - 36 < 0, or β2<36\beta^2 < 36. Taking the square root of both sides gives ∣β∣<6|\beta| < 6, which is equivalent to the interval −6<β<6-6 < \beta < 6.

Question 16

The general solution y(t)y(t) to the differential equation y′′+αy′+4y=0y'' + \alpha y' + 4y = 0 is known to satisfy lim⁡t→∞y(t)=0\lim_{t \to \infty} y(t) = 0. What must be true about the real parameter α\alpha?

  1. α>0\alpha > 0 (correct answer)
  2. α<0\alpha < 0
  3. α=0\alpha = 0
  4. −4<α<4-4 < \alpha < 4
Explanation: For the solutions to decay to zero as t→∞t \to \infty, the real parts of the roots of the characteristic equation must be negative. The characteristic equation is r2+αr+4=0r^2 + \alpha r + 4 = 0. The roots are r=−α±α2−162r = \frac{-\alpha \pm \sqrt{\alpha^2 - 16}}{2}. Case 1: Complex roots (α2−16<0\alpha^2 - 16 < 0). The roots are r=−α/2±i16−α22r = -\alpha/2 \pm i\frac{\sqrt{16-\alpha^2}}{2}. The real part is −α/2-\alpha/2. For this to be negative, we need −α/2<0-\alpha/2 < 0, which implies α>0\alpha > 0. Case 2: Real roots (α2−16≥0\alpha^2 - 16 \ge 0). Both roots must be negative. The sum of the roots is −b/a=−α-b/a = -\alpha, and the product is c/a=4c/a = 4. Since the product is positive, the roots must have the same sign. For them to be negative, their sum must be negative. So, −α<0-\alpha < 0, which implies α>0\alpha > 0. In both cases, α\alpha must be positive.

Question 17

For a particular value of kk, the differential equation y′′+6y′+(k+5)y=0y'' + 6y' + (k+5)y = 0 has a characteristic equation with a repeated real root. What is the general solution for this value of kk?

  1. y(t)=c1e−4t+c2te−4ty(t) = c_1 e^{-4t} + c_2 t e^{-4t}
  2. y(t)=c1e3t+c2te3ty(t) = c_1 e^{3t} + c_2 t e^{3t}
  3. y(t)=c1e−3ty(t) = c_1 e^{-3t}
  4. y(t)=c1e−3t+c2te−3ty(t) = c_1 e^{-3t} + c_2 t e^{-3t} (correct answer)
Explanation: First, find the value of kk. A repeated real root occurs when the discriminant of the characteristic equation is zero. The characteristic equation is r2+6r+(k+5)=0r^2 + 6r + (k+5) = 0. The discriminant is b2−4ac=62−4(1)(k+5)b^2 - 4ac = 6^2 - 4(1)(k+5). Setting this to zero: 36−4k−20=0  ⟹  16−4k=0  ⟹  k=436 - 4k - 20 = 0 \implies 16 - 4k = 0 \implies k=4. Now, substitute k=4k=4 into the ODE: y′′+6y′+9y=0y'' + 6y' + 9y = 0. The characteristic equation is r2+6r+9=0r^2 + 6r + 9 = 0, which is (r+3)2=0(r+3)^2 = 0. This gives a repeated root of r=−3r=-3. The general solution for a repeated root rr is y(t)=c1ert+c2terty(t) = c_1 e^{rt} + c_2 t e^{rt}. Therefore, the solution is y(t)=c1e−3t+c2te−3ty(t) = c_1 e^{-3t} + c_2 t e^{-3t}.

Question 18

A solution to y′′+by′+cy=0y'' + by' + cy = 0 is y(t)=5te−2ty(t) = 5te^{-2t}. What must be the form of the characteristic equation?

  1. r2−4r+4=0r^2 - 4r + 4 = 0
  2. r2+4=0r^2 + 4 = 0
  3. r2+2r=0r^2 + 2r = 0
  4. r2+4r+4=0r^2 + 4r + 4 = 0 (correct answer)
Explanation: When you encounter a linear homogeneous differential equation with a given solution, you need to work backwards from the solution to find the characteristic equation. The key insight is recognizing what type of solution you're dealing with. The given solution y(t)=5te−2ty(t) = 5te^{-2t} has the form tertte^{rt} where r=−2r = -2. This is crucial because solutions of the form tertte^{rt} only arise when the characteristic equation has a repeated root at rr. This happens when the discriminant of the characteristic equation equals zero. For a repeated root at r=−2r = -2, the characteristic equation must be (r−(−2))2=0(r - (-2))^2 = 0, which expands to (r+2)2=0(r + 2)^2 = 0. Expanding this gives us r2+4r+4=0r^2 + 4r + 4 = 0, confirming that answer D is correct. Let's examine why the other options fail: Option A gives r2−4r+4=(r−2)2=0r^2 - 4r + 4 = (r-2)^2 = 0, which has a repeated root at r=2r = 2, producing solutions involving e2te^{2t}, not e−2te^{-2t}. Option B gives r2+4=0r^2 + 4 = 0, yielding complex roots r=±2ir = \pm 2i, which generate trigonometric solutions like cos⁡(2t)\cos(2t) and sin⁡(2t)\sin(2t). Option C gives r2+2r=r(r+2)=0r^2 + 2r = r(r+2) = 0, producing distinct roots r=0r = 0 and r=−2r = -2, which would give solutions e0t=1e^{0t} = 1 and e−2te^{-2t}, but no factor of tt. Study tip: Remember that tertte^{rt} solutions always indicate repeated roots in the characteristic equation. When you see a tt multiplying an exponential, look for (r−exponent)2=0(r - \text{exponent})^2 = 0.

Question 19

The characteristic equation r2+βr+γ=0r^2 + \beta r + \gamma = 0 has discriminant Δ=8\Delta = 8. If one root is r1=3+2r_1 = 3 + \sqrt{2}, what is the value of βγ\beta \gamma?

  1. βγ=49\beta \gamma = 49
  2. βγ=−49\beta \gamma = -49
  3. βγ=42\beta \gamma = 42
  4. βγ=−42\beta \gamma = -42 (correct answer)
Explanation: When you encounter characteristic equations with given roots and discriminants, you're working with the fundamental relationships between a quadratic's coefficients and its solutions. The key insight is using Vieta's formulas along with the discriminant formula. Since the discriminant Δ=β2−4γ=8\Delta = \beta^2 - 4\gamma = 8, and one root is r1=3+2r_1 = 3 + \sqrt{2}, you can find the second root using the discriminant. For a quadratic r2+βr+γ=0r^2 + \beta r + \gamma = 0, if r1=3+2r_1 = 3 + \sqrt{2}, then r2=3−2r_2 = 3 - \sqrt{2} (the roots are −β±Δ2=−β±222\frac{-\beta \pm \sqrt{\Delta}}{2} = \frac{-\beta \pm 2\sqrt{2}}{2}). Now apply Vieta's formulas:
  • Sum of roots: r1+r2=−βr_1 + r_2 = -\beta, so (3+2)+(3−2)=6=−β(3 + \sqrt{2}) + (3 - \sqrt{2}) = 6 = -\beta, giving β=−6\beta = -6
  • Product of roots: r1⋅r2=γr_1 \cdot r_2 = \gamma, so (3+2)(3−2)=9−2=7=γ(3 + \sqrt{2})(3 - \sqrt{2}) = 9 - 2 = 7 = \gamma
Therefore, βγ=(−6)(7)=−42\beta\gamma = (-6)(7) = -42, confirming answer D. Choice A (βγ=49\beta\gamma = 49) likely comes from incorrectly calculating (−6)×7=42(-6) \times 7 = 42 then squaring something. Choice B (βγ=−49\beta\gamma = -49) might result from confusing the sum and product relationships. Choice C (βγ=42\beta\gamma = 42) represents the common error of forgetting the negative sign when β=−6\beta = -6. Remember: always verify your coefficients satisfy the original discriminant condition. Here, (−6)2−4(7)=36−28=8(-6)^2 - 4(7) = 36 - 28 = 8 ✓.

Question 20

A second-order linear ODE has characteristic equation r2−2mr+(m2+n2)=0r^2 - 2mr + (m^2 + n^2) = 0 where mm and nn are real constants with n≠0n \neq 0. What is the general form of the solution?

  1. y(x)=c1e(m+n)x+c2e(m−n)xy(x) = c_1 e^{(m+n)x} + c_2 e^{(m-n)x}
  2. y(x)=emx(c1cos⁡(nx)+c2sin⁡(nx))y(x) = e^{mx}(c_1 \cos(nx) + c_2 \sin(nx)) (correct answer)
  3. y(x)=emx(c1+c2x)y(x) = e^{mx}(c_1 + c_2 x)
  4. y(x)=c1emx+c2e−mxy(x) = c_1 e^{mx} + c_2 e^{-mx}
Explanation: When you encounter a characteristic equation for a second-order linear ODE, the nature of its roots determines the form of the general solution. The key is analyzing the discriminant to classify the roots as real and distinct, real and repeated, or complex conjugates. For the characteristic equation r2−2mr+(m2+n2)=0r^2 - 2mr + (m^2 + n^2) = 0, let's find the discriminant: Δ=(−2m)2−4(1)(m2+n2)=4m2−4m2−4n2=−4n2\Delta = (-2m)^2 - 4(1)(m^2 + n^2) = 4m^2 - 4m^2 - 4n^2 = -4n^2. Since n≠0n \neq 0, we have Δ<0\Delta < 0, indicating complex conjugate roots. Using the quadratic formula: r=2m±−4n22=2m±2ni2=m±nir = \frac{2m \pm \sqrt{-4n^2}}{2} = \frac{2m \pm 2ni}{2} = m \pm ni. When the characteristic equation has complex roots α±βi\alpha \pm \beta i, the general solution is y(x)=eαx(c1cos⁡(βx)+c2sin⁡(βx))y(x) = e^{\alpha x}(c_1 \cos(\beta x) + c_2 \sin(\beta x)). With α=m\alpha = m and β=n\beta = n, this gives us choice B. Choice A represents the solution for real, distinct roots m+nm+n and m−nm-n, which would require a positive discriminant. Choice C is the form for repeated real roots, occurring when the discriminant equals zero. Choice D suggests roots mm and −m-m, which doesn't match our characteristic equation structure. Study tip: Always compute the discriminant first when solving characteristic equations. Negative discriminant means complex roots, leading to exponential-trigonometric solutions. Remember the pattern: ereal part⋅x(cos⁡+sin⁡ of imaginary part)e^{\text{real part} \cdot x}(\cos + \sin \text{ of imaginary part}).