Distinct Real Roots - Differential Equations

Consider the second-order linear homogeneous differential equation $ay'' + by' + cy = 0$ where $a$ , $b$ , and $c$ are constants with $a \neq 0$ . If the characteristic equation has roots $r_1 = -2$ and $r_2 = 3$ , and the solution satisfies $y(0) = 4$ and $y'(0) = -1$ , what is the value of $y(1)$ ?
Answer: $y(1) = 3e^{-2} + e^3$ With distinct real roots r₁ = -2 and r₂ = 3, the general solution is y(t) = c₁ $e^{-2t}$ + c₂ $e^{3t}$ . From y(0) = 4: c₁ + c₂ = 4. Taking the derivative: y'(t) = -2c₁ $e^{-2t}$ + 3c₂ $e^{3t}$ . 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.
Consider the second-order equation $y'' + 2y' - 8y = 0$ . If $u(x) = y(-x)$ where $y(x)$ is a solution to the original equation, which differential equation does $u(x)$ satisfy?
Answer: $u'' - 2u' - 8u = 0$ 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.
Consider two linearly independent solutions $y_1(x)$ and $y_2(x)$ to $y'' + py' + qy = 0$ where the characteristic roots are $r_1 = 2$ and $r_2 = -5$ . If $v(x) = y_1(x) + 2y_2(x)$ and $w(x) = 3y_1(x) - y_2(x)$ , what is the Wronskian $W[v,w](0)$ if $y_1(0) = 1$ , $y_1'(0) = -1$ , $y_2(0) = 2$ , and $y_2'(0) = 3$ ?
Answer: $W[v,w](0) = -35$ 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: - $v(0) = y_1(0) + 2y_2(0) = 1 + 2(2) = 5$ - $v'(0) = y_1'(0) + 2y_2'(0) = -1 + 2(3) = 5$ - $w(0) = 3y_1(0) - y_2(0) = 3(1) - 2 = 1$ - $w'(0) = 3y_1'(0) - y_2'(0) = 3(-1) - 3 = -6$ Therefore: $W[v,w](0) = v(0)w'(0) - v'(0)w(0) = 5(-6) - 5(1) = -30 - 5 = -35$ Choice A (35) represents taking the absolute value or making a sign error. Choice B (-21) likely comes from computational mistakes in the linear combinations. Choice C (21) combines both the wrong magnitude and wrong sign, possibly from misapplying the Wronskian formula. Study tip: Always double-check your arithmetic when computing linear combinations, and remember the Wronskian formula has a subtraction that affects the final sign. Practice computing derivatives of linear combinations until it becomes automatic.
A second-order linear homogeneous differential equation has characteristic polynomial $\lambda^2 - k\lambda - 12 = 0$ where $k > 0$ . If one solution to the differential equation is $y_1(x) = e^{4x}$ , which of the following represents the general solution?
Answer: $y(x) = c_1 e^{4x} + c_2 e^{-3x}$ 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) = c₁ $e^{4x}$ + c₂ $e^{-3x}$ . Choice A incorrectly includes a repeated root term. Choice C uses the wrong second root. Choice D assumes the roots are ±4.
Consider the differential equation $y'' - (a+b)y' + aby = 0$ where $a$ and $b$ are distinct real constants. A particular solution has the property that $\lim_{x \to \infty} y(x) = 0$ and $y(0) = 1$ . If $a = 2$ and $b = -3$ , which condition must the solution satisfy?
Answer: The coefficient of $e^{2x}$ must equal zero, requiring $y'(0) = -3$ 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.
The function $y(x) = 3e^{2x} - 5e^{-x}$ satisfies a second-order linear homogeneous differential equation with constant coefficients. If $z(x)$ is another solution to the same equation with $z(0) = 1$ and $z'(0) = 0$ , what is $z(x)$ ?
Answer: $z(x) = \frac{1}{3}e^{2x} + \frac{2}{3}e^{-x}$ 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) = c₁ $e^{2x}$ + c₂ $e^{-x}$ . For z(x) with z(0) = 1: c₁ + c₂ = 1. Taking z'(x) = 2c₁ $e^{2x}$ - c₂ $e^{-x}$ , 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.

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