Differential Equations Quiz: Distinct Real Roots

What this quiz covers

This quiz focuses on Distinct Real Roots, 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.

Question 1

1. $z(x) = \frac{5}{3}e^{2x} - \frac{2}{3}e^{-x}$
2. $z(x) = \frac{2}{3}e^{2x} + \frac{1}{3}e^{-x}$
3. $z(x) = \frac{1}{3}e^{2x} + \frac{2}{3}e^{-x}$ (correct answer)
4. $z(x) = -\frac{2}{3}e^{2x} + \frac{5}{3}e^{-x}$

Explanation: Since y(x) = 3e^(2x) - 5e^(-x) is a solution, the characteristic equation has roots r₁ = 2 and r₂ = -1. The general solution is y(x) = c₁e^(2x) + c₂e^(-x). For z(x) with z(0) = 1: c₁ + c₂ = 1. Taking z'(x) = 2c₁e^(2x) - c₂e^(-x), and using z'(0) = 0: 2c₁ - c₂ = 0, so c₂ = 2c₁. Substituting: c₁ + 2c₁ = 1, so c₁ = 1/3 and c₂ = 2/3. Therefore z(x) = (1/3)e^(2x) + (2/3)e^(-x). Choice A has coefficients that don't satisfy z(0) = 1. Choice B gives z'(0) = 1/3 ≠ 0. Choice D gives z(0) = 1 but z'(0) = -8/3 ≠ 0.

Question 2

1. $y(1) = 3e^{-2} + e^3$ (correct answer)
2. $y(1) = e^{-2} + 3e^3$
3. $y(1) = 5e^{-2} - e^3$
4. $y(1) = -e^{-2} + 5e^3$

Explanation: With distinct real roots r₁ = -2 and r₂ = 3, the general solution is y(t) = c₁e^(-2t) + c₂e^(3t). From y(0) = 4: c₁ + c₂ = 4. Taking the derivative: y'(t) = -2c₁e^(-2t) + 3c₂e^(3t). From y'(0) = -1: -2c₁ + 3c₂ = -1. Solving the system: c₁ = 3, c₂ = 1. Therefore y(t) = 3e^(-2t) + e^(3t), so y(1) = 3e^(-2) + e^3. Choice B swaps the coefficients. Choice C results from an error in solving the initial value system. Choice D results from sign errors in applying the initial conditions.

Question 3

1. $W[v,w](0) = 35$
2. $W[v,w](0) = -21$
3. $W[v,w](0) = 21$
4. $W[v,w](0) = -35$ (correct answer)

Explanation: When you encounter Wronskian problems involving linear combinations of solutions, remember that the Wronskian measures linear independence and can be computed directly from the definition using the determinant formula. The Wronskian of two functions $v(x)$ and $w(x)$ is defined as $W[v,w](x) = v(x)w'(x) - v'(x)w(x)$. First, you need to find the values and derivatives of $v$ and $w$ at $x = 0$. Given $v(x) = y_1(x) + 2y_2(x)$ and $w(x) = 3y_1(x) - y_2(x)$, calculate:

• $v(0) = y_1(0) + 2y_2(0) = 1 + 2(2) = 5$
• $v'(0) = y_1'(0) + 2y_2'(0) = -1 + 2(3) = 5$
• $w(0) = 3y_1(0) - y_2(0) = 3(1) - 2 = 1$
• $w'(0) = 3y_1'(0) - y_2'(0) = 3(-1) - 3 = -6$

Question 4

1. $u'' + 2u' - 8u = 0$
2. $u'' - 2u' - 8u = 0$ (correct answer)
3. $u'' + 2u' + 8u = 0$
4. $u'' - 2u' + 8u = 0$

Explanation: Let u(x) = y(-x). Then u'(x) = -y'(-x) and u''(x) = y''(-x). Since y(x) satisfies y'' + 2y' - 8y = 0, we have y''(-x) + 2y'(-x) - 8y(-x) = 0 for all x. Substituting: u''(x) + 2(-u'(x)) - 8u(x) = 0, which simplifies to u'' - 2u' - 8u = 0. Choice A assumes no change occurs. Choice C incorrectly changes the sign of the constant term. Choice D incorrectly changes the sign of both the first derivative and constant terms.

Question 5

1. $y(x) = c_1 e^{4x} + c_2 e^{-3x} + c_3 x e^{4x}$
2. $y(x) = c_1 e^{4x} + c_2 e^{-3x}$ (correct answer)
3. $y(x) = c_1 e^{4x} + c_2 e^{3x}$
4. $y(x) = c_1 e^{4x} + c_2 e^{-4x}$

Explanation: Since y₁(x) = e^(4x) is a solution, λ = 4 must be a root of λ² - kλ - 12 = 0. Substituting: 16 - 4k - 12 = 0, so k = 1. The characteristic equation becomes λ² - λ - 12 = 0, which factors as (λ - 4)(λ + 3) = 0. The roots are λ = 4 and λ = -3, both distinct and real. Therefore the general solution is y(x) = c₁e^(4x) + c₂e^(-3x). Choice A incorrectly includes a repeated root term. Choice C uses the wrong second root. Choice D assumes the roots are ±4.

Question 6

1. The solution approaches zero regardless of initial conditions since both roots are negative
2. The coefficient of $e^{-3x}$ must equal zero, requiring $y'(0) = 2$
3. Both exponential terms must be present with $y'(0) = -1$
4. The coefficient of $e^{2x}$ must equal zero, requiring $y'(0) = -3$ (correct answer)

Explanation: When you encounter a second-order linear homogeneous differential equation with constant coefficients, your first step is finding the characteristic equation and its roots, then applying initial conditions to determine which exponential terms can survive. For $y'' - (a+b)y' + aby = 0$ with $a = 2$ and $b = -3$, you get $y'' + y' - 6y = 0$. The characteristic equation $r^2 + r - 6 = 0$ factors as $(r-2)(r+3) = 0$, giving roots $r = 2$ and $r = -3$. This means the general solution is $y = c_1e^{2x} + c_2e^{-3x}$. Since $\lim_{x \to \infty} y(x) = 0$, the term $e^{2x}$ (which grows exponentially) would make this limit infinite unless its coefficient is zero. Therefore, $c_1 = 0$, leaving $y = c_2e^{-3x}$. From $y(0) = 1$, we get $c_2 = 1$, so $y = e^{-3x}$. Taking the derivative: $y'(x) = -3e^{-3x}$, which gives $y'(0) = -3$. Choice A is wrong because having negative roots doesn't automatically ensure the limit is zero - you need specific initial conditions. Choice B incorrectly identifies which coefficient must be zero and gives the wrong derivative value. Choice C wrongly claims both terms must be present, which contradicts the limit condition. Choice D correctly identifies that $c_1 = 0$ (the coefficient of $e^{2x}$) and that $y'(0) = -3$. Study tip: When dealing with boundary conditions at infinity, always check which exponential terms grow versus decay, then set the growing terms' coefficients to zero first.