Math: First-Order Linear Differential Equations
Solving linear differential equations with integrating factors
Math: First-Order Linear Differential Equations
Solving linear differential equations with integrating factors
Math - Grade 9-12
- 1
Identify whether the differential equation y' + 3y = 6 is first-order linear. If it is, write P(x) and Q(x) from the standard form y' + P(x)y = Q(x).
Compare the equation directly to y' + P(x)y = Q(x).
The equation is first-order linear because it contains y' and y to the first power and has the form y' + P(x)y = Q(x). Here P(x) = 3 and Q(x) = 6. - 2
Solve the differential equation y' + 2y = 0.
The integrating factor is e^(∫2 dx) = e^(2x). Multiplying gives (e^(2x)y)' = 0, so e^(2x)y = C. Therefore, y = Ce^(-2x). - 3
Solve the differential equation y' + 4y = 8.
After multiplying by the integrating factor, the left side becomes the derivative of e^(4x)y.
The integrating factor is e^(∫4 dx) = e^(4x). Multiplying gives (e^(4x)y)' = 8e^(4x). Integrating gives e^(4x)y = 2e^(4x) + C, so y = 2 + Ce^(-4x). - 4
Solve the initial value problem y' + y = e^x, with y(0) = 3.
Find the general solution first, then substitute x = 0 and y = 3.
The integrating factor is e^(∫1 dx) = e^x. Multiplying gives (e^x y)' = e^(2x). Integrating gives e^x y = (1/2)e^(2x) + C, so y = (1/2)e^x + Ce^(-x). Using y(0) = 3 gives 3 = 1/2 + C, so C = 5/2. The solution is y = (1/2)e^x + (5/2)e^(-x). - 5
Put the equation 2y' + 6y = 10x into standard linear form, then identify P(x) and Q(x).
Divide every term by 2 to get y' + 3y = 5x. Therefore, P(x) = 3 and Q(x) = 5x. - 6
Solve the differential equation y' - 3y = 6e^(3x).
The coefficient of y is -3, so the integrating factor uses ∫-3 dx.
The integrating factor is e^(∫-3 dx) = e^(-3x). Multiplying gives (e^(-3x)y)' = 6. Integrating gives e^(-3x)y = 6x + C. Therefore, y = e^(3x)(6x + C). - 7
A slope field represents the differential equation y' + y = 0. Describe the general shape of its solutions and write the general solution.
The solutions are exponential decay curves when C is positive and exponential growth downward when C is negative. Solving y' + y = 0 gives y = Ce^(-x). - 8
Solve the differential equation y' + (2/x)y = x^2 for x > 0.
For x > 0, e^(2 ln x) simplifies to x^2.
The integrating factor is e^(∫2/x dx) = e^(2 ln x) = x^2. Multiplying gives (x^2y)' = x^4. Integrating gives x^2y = x^5/5 + C. Therefore, y = x^3/5 + C/x^2. - 9
Solve the initial value problem y' + (1/x)y = 4x, with y(1) = 5 and x > 0.
The integrating factor is e^(∫1/x dx) = x. Multiplying gives (xy)' = 4x^2. Integrating gives xy = (4/3)x^3 + C, so y = (4/3)x^2 + C/x. Using y(1) = 5 gives 5 = 4/3 + C, so C = 11/3. The solution is y = (4/3)x^2 + 11/(3x). - 10
A tank starts with 100 liters of water containing 20 grams of salt. Brine containing 3 grams of salt per liter enters at 2 liters per minute, and the well-mixed solution leaves at 2 liters per minute. Let S(t) be the grams of salt after t minutes. Write the first-order linear differential equation for S(t), then solve it.
Use rate in minus rate out. The volume stays 100 liters because the inflow and outflow rates are equal.
The salt enters at 6 grams per minute. The salt leaves at a rate of (2/100)S = S/50 grams per minute. The equation is S' = 6 - S/50, or S' + S/50 = 6. The solution is S = 300 + Ce^(-t/50). Since S(0) = 20, C = -280. Therefore, S(t) = 300 - 280e^(-t/50). - 11
The graph of several solutions to y' + 2y = 4 is shown. What horizontal line do all solution curves approach as x increases, and why?
Solve the differential equation or look for the constant solution where y' = 0.
All solution curves approach the horizontal line y = 2. The general solution is y = 2 + Ce^(-2x), and the exponential term Ce^(-2x) approaches 0 as x increases. - 12
Explain the error in this work: To solve y' + 5y = 10, a student writes the integrating factor as 5e^x and then multiplies the equation by 5e^x.
The error is that the integrating factor is not found by multiplying the coefficient 5 by e^x. The correct integrating factor is e^(∫5 dx) = e^(5x). Multiplying by e^(5x) makes the left side (e^(5x)y)'.