Repeated Roots - Differential Equations

1. A mass-spring system is governed by the differential equation $m\frac{d^2x}{dt^2} + 2\gamma m\frac{dx}{dt} + kx = 0$ where $m > 0$, $\gamma > 0$, and $k > 0$. If the system is critically damped (repeated roots), and the solution with initial conditions $x(0) = x_0$ and $x'(0) = v_0$ can be written as $x(t) = (A + Bt)e^{-\omega t}$, which relationship must hold between the parameters?
   Answer: $\gamma^2 = \frac{k}{m}$ and $\omega = \sqrt{\frac{k}{m}}$ Dividing the equation by $m$: $\frac{d^2x}{dt^2} + 2\gamma\frac{dx}{dt} + \frac{k}{m}x = 0$. The characteristic equation is $r^2 + 2\gamma r + \frac{k}{m} = 0$. For critical damping (repeated roots), the discriminant must be zero: $(2\gamma)^2 - 4 \cdot 1 \cdot \frac{k}{m} = 0$, which gives $4\gamma^2 = \frac{4k}{m}$, or $\gamma^2 = \frac{k}{m}$. The repeated root is $r = -\gamma$, so $\omega = \gamma = \sqrt{\frac{k}{m}}$.
2. A second-order linear homogeneous differential equation has characteristic polynomial $P(r) = r^2 + pr + q$. If the equation has a repeated root and one particular solution is $y_1(x) = xe^{-2x}$, what is the value of $p + q$?
   Answer: $p + q = 8$ Since $y_1(x) = xe^{-2x}$ is a solution and we have a repeated root, the general solution must be of the form $y = (c_1 + c_2 x)e^{-2x}$. This means the repeated root is $r = -2$. The characteristic polynomial with repeated root $r = -2$ is $(r + 2)^2 = r^2 + 4r + 4$. Therefore $p = 4$ and $q = 4$, giving $p + q = 8$.
3. The differential equation $y'' + ay' + by = 0$ has characteristic equation with a repeated root at $r = -3$. If $u(x) = y(x)e^{3x}$ where $y(x)$ is a solution to the original equation, which differential equation does $u(x)$ satisfy?
   Answer: $u'' = 0$ Since the characteristic equation has repeated root $r = -3$, we have $(r + 3)^2 = r^2 + 6r + 9 = 0$, so $a = 6$ and $b = 9$. The original equation is $y'' + 6y' + 9y = 0$. The general solution is $y = (c_1 + c_2 x)e^{-3x}$. If $u(x) = y(x)e^{3x}$, then $y(x) = u(x)e^{-3x}$. Substituting: $y' = u'e^{-3x} - 3ue^{-3x}$ and $y'' = u''e^{-3x} - 6u'e^{-3x} + 9ue^{-3x}$. Substituting into the original equation: $(u''e^{-3x} - 6u'e^{-3x} + 9ue^{-3x}) + 6(u'e^{-3x} - 3ue^{-3x}) + 9(ue^{-3x}) = 0$. Factoring out $e^{-3x}$: $u'' - 6u' + 9u + 6u' - 18u + 9u = 0$, which simplifies to $u'' = 0$.
4. Consider the equation $y'' - 6y' + 9y = 0$ with general solution $y = (c_1 + c_2 x)e^{3x}$. If $f(x) = y(x) \cdot g(x)$ where $g(x)$ is a twice-differentiable function, and we want $f(x)$ to satisfy $f'' - 6f' + 9f = h(x)$ for some function $h(x)$, what must be true about $g(x)$?
   Answer: $g(x)$ can be any twice-differentiable function This question tests your understanding of how differential operators behave when applied to products of functions. The key insight is that we're multiplying the general solution of a homogeneous equation by another function and asking what constraints this places on that function. Let's work through what happens when we substitute $f(x) = y(x) \cdot g(x)$ into the differential equation $f'' - 6f' + 9f = h(x)$. Using the product rule repeatedly to find $f'$ and $f''$, then substituting into our equation, we get: $[y''g + 2y'g' + yg''] - 6[y'g + yg'] + 9[yg] = h(x)$ Rearranging terms: $g(y'' - 6y' + 9y) + g'(2y' - 6y) + g''y = h(x)$ Since $y$ satisfies $y'' - 6y' + 9y = 0$, the first term vanishes. From the given general solution $y = (c_1 + c_2x)e^{3x}$, we can verify that $2y' - 6y = 0$ as well (this is a special property of repeated roots). Therefore, our equation simplifies to $g''y = h(x)$, which means $g'' = \frac{h(x)}{y}$. This shows that $g(x)$ can indeed be any twice-differentiable function - we simply define $h(x) = g''(x) \cdot y(x)$ and the equation is satisfied. Answer choices A, B, and C all impose unnecessary restrictions on $g(x)$. The beauty of this problem is that no constraints are needed beyond twice-differentiability. Study tip: When dealing with products in differential equations, remember that the differential operator distributes according to product rule patterns, often revealing hidden relationships between the component functions.
5. Consider the differential equation $y'' + 4y' + 4y = 0$. If $z(x) = \int_0^x y(t) dt$ where $y(x)$ is the solution satisfying $y(0) = 0$ and $y'(0) = 1$, what differential equation does $z(x)$ satisfy?
   Answer: $z''' + 4z'' + 4z' = 0$ with $z(0) = z'(0) = 0$, $z''(0) = 1$ Since $z(x) = \int_0^x y(t) dt$, we have $z'(x) = y(x)$ and $z''(x) = y'(x)$. From the original equation $y'' + 4y' + 4y = 0$, substituting: $z''' + 4z'' + 4z' = 0$. For initial conditions: $z(0) = \int_0^0 y(t) dt = 0$, $z'(0) = y(0) = 0$, and $z''(0) = y'(0) = 1$.
6. The functions $y_1(x) = e^{3x}$ and $y_2(x) = xe^{3x}$ are solutions to a second-order linear homogeneous differential equation. If $W(x)$ denotes the Wronskian of these solutions, what is $\frac{W(1)}{W(0)}$?
   Answer: $\frac{W(1)}{W(0)} = e^6$ When you encounter Wronskian problems involving solutions to linear homogeneous differential equations, remember that the Wronskian has a special exponential behavior that often provides shortcuts to avoid tedious calculations. The Wronskian of $y_1(x) = e^{3x}$ and $y_2(x) = xe^{3x}$ is: $W(x) = \begin{vmatrix} e^{3x} & xe^{3x} \\ 3e^{3x} & e^{3x} + 3xe^{3x} \end{vmatrix}$ Computing this determinant: $W(x) = e^{3x}(e^{3x} + 3xe^{3x}) - xe^{3x}(3e^{3x}) = e^{6x}$ Therefore: $\frac{W(1)}{W(0)} = \frac{e^6}{e^0} = e^6$ Answer A gives $e^3$, which would result from incorrectly thinking the Wronskian grows like $e^{3x}$ instead of $e^{6x}$. This mistake often comes from focusing only on the exponential factor in the original solutions rather than properly computing the determinant. Answer B suggests the ratio equals 1, implying the Wronskian is constant. This would only be true if the coefficient function $p(x)$ in the differential equation $y'' + p(x)y' + q(x)y = 0$ were zero, which isn't the case here. Answer C gives $e^9$, which might come from incorrectly squaring the $e^{3x}$ factor and then cubing it, representing a computational error in determinant evaluation. Study tip: For Wronskians of exponential-type solutions, look for patterns in the exponents. The Wronskian often has an exponential form that's predictable from the original solutions' structure, and Abel's theorem can provide useful shortcuts.

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