Calculus: Differential Equations
Solving and interpreting first-order and second-order differential equations
Calculus: Differential Equations
Solving and interpreting first-order and second-order differential equations
Calculus - Grade advanced
- 1
Solve the separable differential equation dy/dx = 3x^2 y, where y > 0.
Move all y terms to one side and all x terms to the other side before integrating.
Separating variables gives (1/y) dy = 3x^2 dx. Integrating gives ln y = x^3 + C, so the general solution is y = Ce^(x^3), where C is a positive constant. - 2
Find the particular solution to dy/dx = 2x(y + 1) with initial condition y(0) = 3.
Separate variables to get dy/(y + 1) = 2x dx. Integrating gives ln|y + 1| = x^2 + C. Since y(0) = 3, we have ln 4 = C, so y + 1 = 4e^(x^2). Therefore, y = 4e^(x^2) - 1. - 3
Solve the linear differential equation dy/dx + 2y = e^x.
Use the integrating factor e^(∫2 dx).
The integrating factor is e^(2x). Multiplying through gives d/dx(e^(2x)y) = e^(3x). Integrating gives e^(2x)y = (1/3)e^(3x) + C, so y = (1/3)e^x + Ce^(-2x). - 4
Solve the initial value problem dy/dx - y = x, with y(0) = 2.
The integrating factor is e^(-x). Multiplying gives d/dx(e^(-x)y) = xe^(-x). Integrating gives e^(-x)y = -(x + 1)e^(-x) + C, so y = -x - 1 + Ce^x. Using y(0) = 2 gives C = 3. The solution is y = -x - 1 + 3e^x. - 5
A population P satisfies the exponential growth model dP/dt = 0.08P. If P(0) = 500, find P(t) and determine the population after 10 years.
Use the standard exponential model P(t) = P0e^(kt).
The solution to dP/dt = 0.08P is P(t) = Ce^(0.08t). Since P(0) = 500, C = 500. Thus P(t) = 500e^(0.08t). After 10 years, P(10) = 500e^0.8, which is approximately 1113 people. - 6
A cooling object follows Newton's law of cooling: dT/dt = -0.2(T - 20), where T is measured in degrees Celsius and t is measured in minutes. If T(0) = 80, find T(t).
The equilibrium temperature is 20 degrees Celsius. Solving gives T - 20 = Ce^(-0.2t). Since T(0) = 80, C = 60. Therefore, T(t) = 20 + 60e^(-0.2t). - 7
For the logistic equation dP/dt = 0.4P(1 - P/1000), identify the carrying capacity and determine whether P is increasing or decreasing when P = 300.
Look for the value of P that makes dP/dt equal to zero.
The carrying capacity is 1000 because the factor 1 - P/1000 becomes zero at P = 1000. When P = 300, both P and 1 - P/1000 are positive, so dP/dt is positive. Therefore, the population is increasing. - 8
Solve the logistic differential equation dP/dt = 0.5P(1 - P/200) with P(0) = 50.
The logistic solution has the form P(t) = 200/(1 + Ae^(-0.5t)). Using P(0) = 50 gives 50 = 200/(1 + A), so 1 + A = 4 and A = 3. Therefore, P(t) = 200/(1 + 3e^(-0.5t)). - 9
Use the slope field for dy/dx = x - y to estimate the sign of the slope at the point (2, 5), and explain what that means for a solution curve passing through that point.
Substitute x = 2 and y = 5 into the differential equation.
At (2, 5), the slope is dy/dx = 2 - 5 = -3. The slope is negative, so a solution curve passing through that point is decreasing there and has a fairly steep downward direction. - 10
Sketch or describe the solution curve through (0, 1) for the differential equation dy/dx = y. What is the exact solution?
The solution curve through (0, 1) is an increasing exponential curve. Since dy/dx = y, the general solution is y = Ce^x. Using y(0) = 1 gives C = 1, so the exact solution is y = e^x. - 11
Solve the second-order differential equation y'' - 5y' + 6y = 0.
Write and factor the characteristic equation.
The characteristic equation is r^2 - 5r + 6 = 0. Factoring gives (r - 2)(r - 3) = 0, so r = 2 and r = 3. The general solution is y = C1e^(2x) + C2e^(3x). - 12
Solve the second-order differential equation y'' + 4y = 0.
The characteristic equation is r^2 + 4 = 0, so r = ±2i. For imaginary roots ±bi, the solution is y = C1cos(bx) + C2sin(bx). Therefore, y = C1cos(2x) + C2sin(2x). - 13
Solve y'' + 2y' + 5y = 0.
Complex roots a ± bi lead to e^(ax)(C1cos bx + C2sin bx).
The characteristic equation is r^2 + 2r + 5 = 0. Using the quadratic formula gives r = -1 ± 2i. Therefore, the general solution is y = e^(-x)(C1cos(2x) + C2sin(2x)). - 14
A mass-spring system is modeled by x'' + 9x = 0, where x is displacement from equilibrium. If x(0) = 2 and x'(0) = 0, find x(t).
The general solution is x(t) = C1cos(3t) + C2sin(3t). Since x(0) = 2, C1 = 2. The derivative is x'(t) = -3C1sin(3t) + 3C2cos(3t). Since x'(0) = 0, 3C2 = 0, so C2 = 0. Therefore, x(t) = 2cos(3t). - 15
Match the phase line behavior for dP/dt = P(4 - P): identify the equilibrium solutions and classify each as stable or unstable.
Test one value of P in each interval separated by the equilibrium points.
The equilibrium solutions occur when P(4 - P) = 0, so P = 0 and P = 4. For P < 0, dP/dt is negative. For 0 < P < 4, dP/dt is positive. For P > 4, dP/dt is negative. Solutions move away from P = 0, so P = 0 is unstable. Solutions move toward P = 4 from both sides, so P = 4 is stable.