Characteristic Equation - Differential Equations

The characteristic equation $r^2 - 6r + k = 0$ corresponds to a second-order linear ODE. For which value of $k$ will the general solution contain terms of the form $e^{3x}(A\cos(2x) + B\sin(2x))$ ?
Answer: $k = 13$ The given solution form $e^{3x}(A\cos(2x) + B\sin(2x))$ indicates complex conjugate roots of the form $3 \pm 2i$ . Using the quadratic formula on $r^2 - 6r + k = 0$ : $r = \frac{6 \pm \sqrt{36 - 4k}}{2} = 3 \pm \frac{\sqrt{36 - 4k}}{2}$ . For roots $3 \pm 2i$ , we need $\frac{\sqrt{36 - 4k}}{2} = 2i$ , which means $36 - 4k = -16$ , so $k = 13$ . Choice A gives real distinct roots, choice B gives repeated real roots, and choice D gives complex roots with wrong imaginary part.
The general solution of a second-order linear homogeneous ODE is given by $y(x) = c_1 e^{-2x} + c_2 xe^{-2x}$ . If this same ODE is written in the standard form $y'' + py' + qy = 0$ , what is the value of the discriminant of its characteristic equation?
Answer: The discriminant equals $p^2 - 4q = 0$ The solution $y(x) = c_1 e^{-2x} + c_2 xe^{-2x}$ indicates a repeated root $r = -2$ in the characteristic equation $r^2 + pr + q = 0$ . For a repeated root, the characteristic equation must be $(r + 2)^2 = r^2 + 4r + 4 = 0$ , so $p = 4$ and $q = 4$ . The discriminant is $p^2 - 4q = 16 - 16 = 0$ , which confirms the repeated root case. Choices A and D suggest distinct real roots, while choice C suggests complex roots.
The function $y_1(t) = e^{2t}\sin(5t)$ 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?
Answer: $y(t) = e^{2t}\cos(5t)$ 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 $e^{2t}\sin(5t)$ arises from characteristic roots $r = 2 \pm 5i$ . The general solution is of the form $y(t) = e^{2t}(c_1\cos(5t) + c_2\sin(5t))$ . Since $y_1(t) = e^{2t}\sin(5t)$ is a solution (corresponding to $c_1=0, c_2=1$ ), the other linearly independent basis function, $y_2(t) = e^{2t}\cos(5t)$ , must also be a solution (corresponding to $c_1=1, c_2=0$ ).
The differential equation $y'' - 10y' + c y = 0$ has a general solution of the form $y(t) = c_1 e^{r_1 t} + c_2 e^{r_2 t}$ where $r_1$ and $r_2$ are distinct real numbers. If one root of the characteristic equation is $r_1 = 4$ , what is the general solution?
Answer: $y(t) = c_1 e^{4t} + c_2 e^{6t}$ When you encounter a second-order linear homogeneous differential equation like $y'' - 10y' + cy = 0$ , you solve it by finding the characteristic equation and its roots. The characteristic equation is formed by substituting $y = e^{rt}$ , giving you $r^2 - 10r + c = 0$ . Since you're told that one root is $r_1 = 4$ and the roots are distinct real numbers, you can find the second root using the relationship between roots and coefficients. For a quadratic $r^2 - 10r + c = 0$ , the sum of roots equals 10 (the negative of the coefficient of $r$ ). Therefore: $r_1 + r_2 = 10$ , so $4 + r_2 = 10$ , which gives $r_2 = 6$ . The general solution is $y(t) = c_1 e^{4t} + c_2 e^{6t}$ , which is answer C. Let's examine why the other options are incorrect: A) $y(t) = c_1 e^{4t} + c_2 t e^{4t}$ represents the case of repeated roots, not distinct roots as specified in the problem. B) $y(t) = c_1 e^{4t} + c_2 e^{-14t}$ would require roots that sum to $4 + (-14) = -10$ , but our coefficient pattern requires a sum of +10. D) $y(t) = c_1 e^{4t} + c_2 e^{5t}$ 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 $r^2 + br + c = 0$ , the sum of roots equals $-b$ and the product equals $c$ . This saves time compared to solving the full quadratic.
Consider the differential equation $y'' + \beta y' + 9y = 0$ . For which values of the real parameter $\beta$ will all non-trivial solutions exhibit oscillatory behavior?
Answer: $-6 < \beta < 6$ 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 $r^2 + \beta r + 9 = 0$ . The roots are complex when the discriminant is negative. The discriminant is $D = b^2 - 4ac = \beta^2 - 4(1)(9) = \beta^2 - 36$ . For complex roots, we need $D < 0$ , which means $\beta^2 - 36 < 0$ , or $\beta^2 < 36$ . Taking the square root of both sides gives $|\beta| < 6$ , which is equivalent to the interval $-6 < \beta < 6$ .
For a particular value of $k$ , the differential equation $y'' + 6y' + (k+5)y = 0$ has a characteristic equation with a repeated real root. What is the general solution for this value of $k$ ?
Answer: $y(t) = c_1 e^{-3t} + c_2 t e^{-3t}$ First, find the value of $k$ . A repeated real root occurs when the discriminant of the characteristic equation is zero. The characteristic equation is $r^2 + 6r + (k+5) = 0$ . The discriminant is $b^2 - 4ac = 6^2 - 4(1)(k+5)$ . Setting this to zero: $36 - 4k - 20 = 0 \implies 16 - 4k = 0 \implies k=4$ . Now, substitute $k=4$ into the ODE: $y'' + 6y' + 9y = 0$ . The characteristic equation is $r^2 + 6r + 9 = 0$ , which is $(r+3)^2 = 0$ . This gives a repeated root of $r=-3$ . The general solution for a repeated root $r$ is $y(t) = c_1 e^{rt} + c_2 t e^{rt}$ . Therefore, the solution is $y(t) = c_1 e^{-3t} + c_2 t e^{-3t}$ .
Consider the equation $2y'' + by' + 8y = 0$ , where $b \ge 0$ . The behavior of the solutions transitions from oscillatory to non-oscillatory as $b$ increases. At what value of $b$ does this transition occur?
Answer: $b=8$ 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 $2r^2 + br + 8 = 0$ . The discriminant is $D = b^2 - 4ac = b^2 - 4(2)(8) = b^2 - 64$ . Setting the discriminant to zero gives $b^2 - 64 = 0$ . Since the problem states $b \ge 0$ , we take the positive root, $b=8$ . For $0 \le b < 8$ , the roots are complex (oscillatory), and for $b > 8$ , the roots are distinct real (non-oscillatory). The transition occurs at $b=8$ .
The motion of a mechanical system is described by $3y'' + 15y = -12y'$ . Which of the following is the general solution $y(t)$ for the system's displacement?
Answer: $y(t) = e^{-2t}(c_1 \cos(t) + c_2 \sin(t))$ First, rewrite the differential equation in the standard form $ay'' + by' + cy = 0$ . Adding $12y'$ to both sides gives $3y'' + 12y' + 15y = 0$ . Divide the entire equation by 3 to simplify: $y'' + 4y' + 5y = 0$ . The characteristic equation is $r^2 + 4r + 5 = 0$ . Using the quadratic formula, the roots are $r = \frac{-4 \pm \sqrt{4^2 - 4(1)(5)}}{2(1)} = \frac{-4 \pm \sqrt{16 - 20}}{2} = \frac{-4 \pm \sqrt{-4}}{2} = \frac{-4 \pm 2i}{2} = -2 \pm i$ . The roots are a complex conjugate pair $\alpha \pm i\beta$ with $\alpha=-2$ and $\beta=1$ . The general solution is $y(t) = e^{\alpha t}(c_1 \cos(\beta t) + c_2 \sin(\beta t))$ , which is $y(t) = e^{-2t}(c_1 \cos(t) + c_2 \sin(t))$ .
The characteristic equation for $ay''+by'+cy=0$ is $ar^2+br+c=0$ . If $a>0$ , $c>0$ , and $b = \sqrt{4ac}$ , which of the following best describes the non-trivial solutions $y(t)$ ?
Answer: Solutions are non-oscillatory and decay to zero as $t \to \infty$ . The condition $b = \sqrt{4ac}$ is equivalent to $b^2 = 4ac$ , which means the discriminant $b^2 - 4ac$ is zero. This indicates a repeated real root for the characteristic equation. The root is given by $r = \frac{-b}{2a}$ . Since $a>0$ and $b=\sqrt{4ac}$ must be positive (as $a,c>0$ ), the root $r = \frac{-b}{2a}$ is negative. The general solution is $y(t) = (c_1 + c_2 t)e^{rt}$ . Since $r<0$ , the exponential term $e^{rt}$ goes to zero as $t \to \infty$ . Although the term $c_2t$ grows, the exponential decay is dominant, so $\lim_{t \to \infty} y(t) = 0$ . The solution is non-oscillatory because the roots are real.
Two linearly independent solutions to $y'' + py' + qy = 0$ are $y_1(t) = e^{at}\cos(bt)$ and $y_2(t) = e^{at}\sin(bt)$ , where $a$ and $b$ are non-zero real numbers. What is the value of $q$ in terms of $a$ and $b$ ?
Answer: $q = a^2 + b^2$ When you encounter a second-order linear homogeneous differential equation with complex exponential solutions like $e^{at}\cos(bt)$ and $e^{at}\sin(bt)$ , you're dealing with a characteristic equation that has complex roots. This form tells you immediately that the characteristic roots are $r = a \pm bi$ . For the differential equation $y'' + py' + qy = 0$ , the characteristic equation is $r^2 + pr + q = 0$ . Since we know the roots are $r_1 = a + bi$ and $r_2 = a - bi$ , we can use Vieta's formulas to find the coefficients. The sum of roots gives us $r_1 + r_2 = (a + bi) + (a - bi) = 2a = -p$ , so $p = -2a$ . The product of roots gives us $r_1 \cdot r_2 = (a + bi)(a - bi) = a^2 - (bi)^2 = a^2 - b^2i^2 = a^2 + b^2 = q$ . Therefore, $q = a^2 + b^2$ , which is answer C. Looking at the wrong answers: A) $q = -2a$ confuses the coefficient $q$ with $p$ ; this is actually what $-p$ equals. B) $q = a^2 - b^2$ makes the common error of forgetting that $i^2 = -1$ when expanding $(a + bi)(a - bi)$ . D) $q = b^2 - a^2$ reverses the correct relationship and also misses the sign change from $i^2$ . Remember: when you see exponential-trigonometric solutions, the characteristic roots are complex conjugates $a \pm bi$ , and their product always gives $a^2 + b^2$ for the constant term.
Let $y_1(t)$ be the solution to $y'' + 9y = 0$ with $y(0)=1, y'(0)=0$ , and let $y_2(t)$ be the solution to $y'' + 6y' + 9y = 0$ with $y(0)=1, y'(0)=0$ . Which statement correctly compares the forms of their characteristic roots?
Answer: The roots for $y_1$ are pure imaginary, while the roots for $y_2$ are a repeated real number. 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 $y = e^{rt}$ and finding the roots of the resulting characteristic equation. For $y_1(t)$ : The equation $y'' + 9y = 0$ gives the characteristic equation $r^2 + 9 = 0$ , so $r^2 = -9$ . This yields $r = \pm 3i$ , which are pure imaginary numbers (no real part, only imaginary components). For $y_2(t)$ : The equation $y'' + 6y' + 9y = 0$ gives $r^2 + 6r + 9 = 0$ . This factors as $(r + 3)^2 = 0$ , yielding the repeated real root $r = -3$ . Looking at the answer choices: Choice A incorrectly reverses which equation has repeated roots and gets the type wrong for $y_1$ . Choice B wrongly claims $y_2$ has distinct real roots when it actually has repeated roots. Choice C incorrectly states $y_1$ has repeated real roots and $y_2$ has complex conjugates. Choice D correctly identifies that $y_1$ has pure imaginary roots ( $\pm 3i$ ) while $y_2$ has a repeated real root ( $-3$ ). 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.
Consider two second-order linear ODEs with characteristic equations $r^2 + ar + b = 0$ and $r^2 + cr + d = 0$ . If the first equation has roots $2 \pm 3i$ and the second has roots $-2 \pm 3i$ , what is the relationship between the coefficients?
Answer: $a + c = 0$ and $b = d$ 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 $r^2 + pr + q = 0$ with roots $r_1$ and $r_2$ , we have $r_1 + r_2 = -p$ and $r_1 \cdot r_2 = q$ . For the first equation with roots $2 \pm 3i$ , the sum is $(2 + 3i) + (2 - 3i) = 4$ , so $a = -4$ . The product is $(2 + 3i)(2 - 3i) = 4 - 9i^2 = 4 + 9 = 13$ , so $b = 13$ . For the second equation with roots $-2 \pm 3i$ , the sum is $(-2 + 3i) + (-2 - 3i) = -4$ , so $c = 4$ . The product is $(-2 + 3i)(-2 - 3i) = 4 - 9i^2 = 13$ , so $d = 13$ . Therefore, $a + c = -4 + 4 = 0$ and $b = d = 13$ . Choice A incorrectly calculates $a - c = -4 - 4 = -8$ , not 4, and while $b - d = 0$ is true, the first condition fails. Choice B claims $a = c$ , but $-4 \neq 4$ , and $b + d = 26$ , not 0. Choice C correctly identifies $a + c = 0$ but incorrectly states $b + d = 26$ instead of recognizing that $b = d$ . 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.
For the differential equation $y'' - 4y' + ky = 0$ , the characteristic equation has two distinct real roots when $k < 4$ , repeated real roots when $k = 4$ , and complex conjugate roots when $k > 4$ . If a particular solution is known to be $y_p(x) = x^2 e^{2x}$ , what can be concluded about the value of $k$ ?
Answer: $k = 4$ and the characteristic equation has a double root at $r = 2$ The solution form $y_p(x) = x^2 e^{2x}$ suggests that $r = 2$ is a repeated root of multiplicity at least 2 (since we have $x^2$ as a factor). For $r^2 - 4r + k = 0$ , if $r = 2$ is a double root, then $(r-2)^2 = r^2 - 4r + 4 = 0$ , so $k = 4$ . 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 $k$ .
The general solution $y(t)$ to the differential equation $y'' + \alpha y' + 4y = 0$ is known to satisfy $\lim_{t \to \infty} y(t) = 0$ . What must be true about the real parameter $\alpha$ ?
Answer: $\alpha > 0$ For the solutions to decay to zero as $t \to \infty$ , the real parts of the roots of the characteristic equation must be negative. The characteristic equation is $r^2 + \alpha r + 4 = 0$ . The roots are $r = \frac{-\alpha \pm \sqrt{\alpha^2 - 16}}{2}$ . Case 1: Complex roots ( $\alpha^2 - 16 < 0$ ). The roots are $r = -\alpha/2 \pm i\frac{\sqrt{16-\alpha^2}}{2}$ . The real part is $-\alpha/2$ . For this to be negative, we need $-\alpha/2 < 0$ , which implies $\alpha > 0$ . Case 2: Real roots ( $\alpha^2 - 16 \ge 0$ ). Both roots must be negative. The sum of the roots is $-b/a = -\alpha$ , and the product is $c/a = 4$ . Since the product is positive, the roots must have the same sign. For them to be negative, their sum must be negative. So, $-\alpha < 0$ , which implies $\alpha > 0$ . In both cases, $\alpha$ must be positive.
A second-order linear ODE has characteristic equation $r^2 - 2mr + (m^2 + n^2) = 0$ where $m$ and $n$ are real constants with $n \neq 0$ . What is the general form of the solution?
Answer: $y(x) = e^{mx}(c_1 \cos(nx) + c_2 \sin(nx))$ 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 $r^2 - 2mr + (m^2 + n^2) = 0$ , let's find the discriminant: $\Delta = (-2m)^2 - 4(1)(m^2 + n^2) = 4m^2 - 4m^2 - 4n^2 = -4n^2$ . Since $n \neq 0$ , we have $\Delta < 0$ , indicating complex conjugate roots. Using the quadratic formula: $r = \frac{2m \pm \sqrt{-4n^2}}{2} = \frac{2m \pm 2ni}{2} = m \pm ni$ . When the characteristic equation has complex roots $\alpha \pm \beta i$ , the general solution is $y(x) = e^{\alpha x}(c_1 \cos(\beta x) + c_2 \sin(\beta x))$ . With $\alpha = m$ and $\beta = n$ , this gives us choice B. Choice A represents the solution for real, distinct roots $m+n$ and $m-n$ , 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 $m$ and $-m$ , 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: $e^{\text{real part} \cdot x}(\cos + \sin \text{ of imaginary part})$ .
The general solution to a constant-coefficient, homogeneous second-order linear differential equation is given by $y(t) = c_1 e^{-3t} + c_2 t e^{-3t}$ . What is the differential equation?
Answer: $y'' + 6y' + 9y = 0$ The form of the general solution $y(t) = c_1 e^{rt} + c_2 t e^{rt}$ indicates a repeated real root $r$ for the characteristic equation. Here, the solution is $y(t) = c_1 e^{-3t} + c_2 t e^{-3t}$ , so the repeated root is $r = -3$ . The characteristic equation must be $(r - (-3))^2 = 0$ , which expands to $(r+3)^2 = 0$ , or $r^2 + 6r + 9 = 0$ . This characteristic equation corresponds to the differential equation $y'' + 6y' + 9y = 0$ .
For the differential equation $4y'' + 4y' + ky = 0$ , determine the value of $k$ such that the characteristic equation has roots with real part equal to $-\frac{1}{2}$ and the general solution contains oscillatory terms.
Answer: $k = 2$ The characteristic equation is $4r^2 + 4r + k = 0$ , or $r^2 + r + \frac{k}{4} = 0$ . Using the quadratic formula: $r = \frac{-1 \pm \sqrt{1 - k}}{2}$ . For oscillatory terms, we need complex roots, so $1 - k < 0$ , meaning $k > 1$ . For real part $-\frac{1}{2}$ , we need $r = -\frac{1}{2} \pm bi$ where $b \neq 0$ . From the formula, the real part is $-\frac{1}{2}$ (which matches) and $b = \frac{\sqrt{k-1}}{2}$ . For oscillatory behavior, we need $k > 1$ . Among the choices greater than 1, we check: if $k = 2$ , then $r = -\frac{1}{2} \pm \frac{i}{2}$ , giving oscillatory solutions. Choice A gives repeated roots, choice B gives the boundary case, and choice D gives a different real part.
The characteristic equation $r^2 + \beta r + \gamma = 0$ has discriminant $\Delta = 8$ . If one root is $r_1 = 3 + \sqrt{2}$ , what is the value of $\beta \gamma$ ?
Answer: $\beta \gamma = -42$ 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 $\Delta = \beta^2 - 4\gamma = 8$ , and one root is $r_1 = 3 + \sqrt{2}$ , you can find the second root using the discriminant. For a quadratic $r^2 + \beta r + \gamma = 0$ , if $r_1 = 3 + \sqrt{2}$ , then $r_2 = 3 - \sqrt{2}$ (the roots are $\frac{-\beta \pm \sqrt{\Delta}}{2} = \frac{-\beta \pm 2\sqrt{2}}{2}$ ). Now apply Vieta's formulas: - Sum of roots: $r_1 + r_2 = -\beta$ , so $(3 + \sqrt{2}) + (3 - \sqrt{2}) = 6 = -\beta$ , giving $\beta = -6$ - Product of roots: $r_1 \cdot r_2 = \gamma$ , so $(3 + \sqrt{2})(3 - \sqrt{2}) = 9 - 2 = 7 = \gamma$ Therefore, $\beta\gamma = (-6)(7) = -42$ , confirming answer D. Choice A ( $\beta\gamma = 49$ ) likely comes from incorrectly calculating $(-6) \times 7 = 42$ then squaring something. Choice B ( $\beta\gamma = -49$ ) might result from confusing the sum and product relationships. Choice C ( $\beta\gamma = 42$ ) represents the common error of forgetting the negative sign when $\beta = -6$ . Remember: always verify your coefficients satisfy the original discriminant condition. Here, $(-6)^2 - 4(7) = 36 - 28 = 8$ ✓.
A solution to $y'' + by' + cy = 0$ is $y(t) = 5te^{-2t}$ . What must be the form of the characteristic equation?
Answer: $r^2 + 4r + 4 = 0$ 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 $y(t) = 5te^{-2t}$ has the form $te^{rt}$ where $r = -2$ . This is crucial because solutions of the form $te^{rt}$ only arise when the characteristic equation has a repeated root at $r$ . This happens when the discriminant of the characteristic equation equals zero. For a repeated root at $r = -2$ , the characteristic equation must be $(r - (-2))^2 = 0$ , which expands to $(r + 2)^2 = 0$ . Expanding this gives us $r^2 + 4r + 4 = 0$ , confirming that answer D is correct. Let's examine why the other options fail: Option A gives $r^2 - 4r + 4 = (r-2)^2 = 0$ , which has a repeated root at $r = 2$ , producing solutions involving $e^{2t}$ , not $e^{-2t}$ . Option B gives $r^2 + 4 = 0$ , yielding complex roots $r = \pm 2i$ , which generate trigonometric solutions like $\cos(2t)$ and $\sin(2t)$ . Option C gives $r^2 + 2r = r(r+2) = 0$ , producing distinct roots $r = 0$ and $r = -2$ , which would give solutions $e^{0t} = 1$ and $e^{-2t}$ , but no factor of $t$ . Study tip: Remember that $te^{rt}$ solutions always indicate repeated roots in the characteristic equation. When you see a $t$ multiplying an exponential, look for $(r - \text{exponent})^2 = 0$ .
Consider the second-order linear homogeneous ODE $ay'' + by' + cy = 0$ where $a$ , $b$ , and $c$ are real constants with $a \neq 0$ . If the characteristic equation has roots $r_1 = 2 + 3i$ and $r_2 = 2 - 3i$ , which of the following must be true about the coefficients?
Answer: $b = -4a$ and $c = 13a$ For a quadratic equation $ar^2 + br + c = 0$ with roots $r_1 = 2 + 3i$ and $r_2 = 2 - 3i$ , we use Vieta's formulas. The sum of roots is $r_1 + r_2 = (2 + 3i) + (2 - 3i) = 4 = -\frac{b}{a}$ , so $b = -4a$ . The product of roots is $r_1 \cdot r_2 = (2 + 3i)(2 - 3i) = 4 - 9i^2 = 4 + 9 = 13 = \frac{c}{a}$ , so $c = 13a$ . Choice B has the wrong sign for $b$ . Choice C has the wrong sign for $c$ . Choice D has wrong signs for both coefficients.

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