Separable First-Order Differential Equations
Solving differential equations by separating variables
Separable First-Order Differential Equations
Solving differential equations by separating variables
Math - Grade 9-12
- 1
Solve the differential equation dy/dx = 3x^2.
This equation is already in a form where you can integrate with respect to x.
Integrating both sides gives y = x^3 + C, where C is an arbitrary constant. - 2
Solve the differential equation dy/dx = 2xy.
Separate the variables to get (1/y) dy = 2x dx. Integrating gives ln|y| = x^2 + C, so y = Ce^(x^2), where C is an arbitrary constant. - 3
Solve the differential equation dy/dx = y/5.
Move all expressions involving y to one side and all expressions involving x to the other side.
Separate the variables to get (1/y) dy = (1/5) dx. Integrating gives ln|y| = x/5 + C, so y = Ce^(x/5), where C is an arbitrary constant. - 4
Find the particular solution to dy/dx = 4x with initial condition y(0) = 7.
Integrating gives y = 2x^2 + C. Using y(0) = 7 gives C = 7, so the particular solution is y = 2x^2 + 7. - 5
Find the particular solution to dy/dx = xy with initial condition y(0) = 3.
After finding the general solution, substitute x = 0 and y = 3.
Separate the variables to get (1/y) dy = x dx. Integrating gives ln|y| = x^2/2 + C, so y = Ce^(x^2/2). Using y(0) = 3 gives C = 3, so y = 3e^(x^2/2). - 6
Solve the differential equation dy/dx = x/(y), assuming y is not 0.
Separate the variables to get y dy = x dx. Integrating gives y^2/2 = x^2/2 + C. Multiplying by 2 gives y^2 = x^2 + C, where C is an arbitrary constant. - 7
Solve the differential equation dy/dx = (x + 1)(y - 2).
Treat y - 2 as the expression that belongs with dy.
Separate the variables to get 1/(y - 2) dy = (x + 1) dx. Integrating gives ln|y - 2| = x^2/2 + x + C, so y = 2 + Ce^(x^2/2 + x). - 8
Find the particular solution to dy/dx = 6x/(y^2) with initial condition y(1) = 2.
Separate the variables to get y^2 dy = 6x dx. Integrating gives y^3/3 = 3x^2 + C, so y^3 = 9x^2 + C. Using y(1) = 2 gives 8 = 9 + C, so C = -1. The particular solution is y^3 = 9x^2 - 1, or y = cuberoot(9x^2 - 1). - 9
A population P grows at a rate proportional to its size: dP/dt = 0.08P. If P(0) = 500, find P(t).
This is exponential growth because the rate is proportional to the current population.
Separate the variables to get (1/P) dP = 0.08 dt. Integrating gives ln|P| = 0.08t + C, so P = Ce^(0.08t). Using P(0) = 500 gives C = 500, so P(t) = 500e^(0.08t). - 10
An object cools according to dT/dt = -0.2(T - 70), where T is the temperature in degrees Fahrenheit and 70 is the room temperature. If T(0) = 150, find T(t).
Separate the variables to get 1/(T - 70) dT = -0.2 dt. Integrating gives ln|T - 70| = -0.2t + C, so T - 70 = Ce^(-0.2t). Using T(0) = 150 gives C = 80, so T(t) = 70 + 80e^(-0.2t). - 11
Determine whether the differential equation dy/dx = x + y is separable. Explain your answer.
A separable equation can be rearranged so all y terms are on one side and all x terms are on the other side.
The equation is not separable because x + y cannot be written as a product of one function of x and one function of y. The variables cannot be separated into the form g(y) dy = f(x) dx. - 12
A slope field for dy/dx = y is shown. Describe how the slopes change as y increases, and state the general solution.
The slope at each point depends only on the y-value.
For dy/dx = y, the slope is larger and positive when y is larger and positive, zero when y = 0, and negative when y is negative. The general solution is y = Ce^x.