Test Questions on Solving Homogeneous 2nd Order ODEs: Real Distinct Roots

📚 Quick Study Guide

• 🔢 A second-order homogeneous ordinary differential equation (ODE) has the form $ay'' + by' + cy = 0$, where $a$, $b$, and $c$ are constants.
• 🔍 To solve, we assume a solution of the form $y = e^{rx}$, where $r$ is a constant.
• 📝 Substituting $y = e^{rx}$ into the ODE, we get the characteristic equation: $ar^2 + br + c = 0$.
• 💡 If the characteristic equation has two distinct real roots $r_1$ and $r_2$, the general solution is given by $y(x) = c_1e^{r_1x} + c_2e^{r_2x}$, where $c_1$ and $c_2$ are arbitrary constants.
• 📌 To find $c_1$ and $c_2$, we use the initial conditions $y(x_0) = y_0$ and $y'(x_0) = y'_0$.

🧪 Practice Quiz

1. What is the general solution to the differential equation $y'' - 5y' + 6y = 0$?
1. $y(x) = c_1e^{2x} + c_2e^{3x}$
2. $y(x) = c_1e^{-2x} + c_2e^{-3x}$
3. $y(x) = c_1e^{2x} + c_2xe^{2x}$
4. $y(x) = c_1e^{3x} + c_2xe^{3x}$
2. Find the solution to $y'' - y' - 2y = 0$ with initial conditions $y(0) = 1$ and $y'(0) = -1$.
1. $y(x) = e^{-x}$
2. $y(x) = e^{2x}$
3. $y(x) = \frac{1}{3}e^{-x} + \frac{2}{3}e^{2x}$
4. $y(x) = \frac{2}{3}e^{-x} + \frac{1}{3}e^{2x}$
3. What is the characteristic equation for the ODE $2y'' + 3y' - 2y = 0$?
1. $2r^2 + 3r - 2 = 0$
2. $2r^2 - 3r + 2 = 0$
3. $r^2 + 3r - 2 = 0$
4. $r^2 - 3r + 2 = 0$
4. The general solution of $y'' - 4y = 0$ is?
1. $y(x) = c_1e^{2x} + c_2e^{-2x}$
2. $y(x) = c_1\cos(2x) + c_2\sin(2x)$
3. $y(x) = c_1e^{4x} + c_2e^{-4x}$
4. $y(x) = c_1e^{2x} + c_2xe^{2x}$
5. Given the ODE $y'' + y' - 6y = 0$, what are the roots of the characteristic equation?
1. $r = 2, -3$
2. $r = -2, 3$
3. $r = 1, -6$
4. $r = -1, 6$
6. Solve $y'' - 3y' + 2y = 0$ with $y(0) = 2$ and $y'(0) = 3$.
1. $y(x) = e^x + e^{2x}$
2. $y(x) = 2e^x - e^{2x}$
3. $y(x) = 3e^x - e^{2x}$
4. $y(x) = 4e^x - 2e^{2x}$
7. What is the solution to $4y'' - 9y = 0$?
1. $y(x) = c_1e^{\frac{3}{2}x} + c_2e^{-\frac{3}{2}x}$
2. $y(x) = c_1\cos(\frac{3}{2}x) + c_2\sin(\frac{3}{2}x)$
3. $y(x) = c_1e^{\frac{9}{4}x} + c_2e^{-\frac{9}{4}x}$
4. $y(x) = c_1e^{\frac{3}{2}x} + c_2xe^{\frac{3}{2}x}$