Welcome to our tutorial on solving explicit differential equations. An explicit differential equation has the form dy/dx equals f of x and y, where the derivative is explicitly expressed in terms of the independent and dependent variables. The solution process involves four main steps: First, identify the type of equation. Second, apply the appropriate solution method. Third, solve for the general solution. And fourth, apply any initial conditions to find a particular solution. On the right, you can see solution curves for the simple differential equation dy/dx equals y, which has the general solution y equals C times e to the x.
Let's look at separable differential equations, which have the form dy/dx equals g of x times h of y. The key characteristic is that we can separate the variables. To solve these equations, we follow three steps: First, separate the variables by rewriting the equation as dy over h of y equals g of x dx. Second, integrate both sides. And third, solve for y in terms of x. Let's consider the example dy/dx equals x times y. This is separable because we can rewrite it as dy over y equals x dx. Integrating both sides gives us the natural log of the absolute value of y equals x squared over 2 plus a constant. Solving for y, we get y equals C times e raised to the power of x squared over 2. The graph shows several solution curves for different values of the constant C.
Now let's examine linear first-order differential equations, which have the form dy/dx plus P of x times y equals Q of x. To solve these equations, we use the integrating factor method. First, find the integrating factor mu of x, which equals e raised to the integral of P of x dx. Second, multiply both sides of the equation by this integrating factor. Third, recognize that the left side becomes the derivative of mu of x times y. Fourth, integrate both sides. And finally, solve for y. Let's work through an example: dy/dx plus y equals e to the x. Here, P of x equals 1, and Q of x equals e to the x. The integrating factor is e to the x. Multiplying both sides by e to the x, we get e to the x times dy/dx plus e to the x times y equals e to the 2x. The left side is the derivative of e to the x times y. Integrating both sides, we get e to the x times y equals e to the 2x over 2 plus C. Solving for y, we get y equals e to the x over 2 plus C times e to the negative x. The red curve shows the particular solution when C equals 0, while the blue curves show the general solution for different values of C.
Now let's discuss Initial Value Problems, or IVPs. An IVP consists of a differential equation together with an initial condition that specifies the value of the dependent variable at a particular point. To solve an IVP, we first find the general solution of the differential equation, which contains an arbitrary constant. Then we apply the initial condition to determine the specific value of this constant. Finally, we substitute this value back into the general solution to obtain the particular solution that satisfies both the differential equation and the initial condition. Let's work through an example: dy/dx equals 2xy with the initial condition y of 0 equals 3. First, we find the general solution. This is a separable equation, so we separate the variables: dy over y equals 2x dx. Integrating both sides gives us the natural log of the absolute value of y equals x squared plus a constant. Solving for y, we get y equals C times e raised to the power of x squared. Next, we apply the initial condition: when x equals 0, y equals 3. Substituting, we get 3 equals C times e to the power of 0 squared, which simplifies to C equals 3. Therefore, our particular solution is y equals 3 times e raised to the power of x squared. On the graph, the blue curves represent the general solution for different values of C, while the red curve shows our particular solution that passes through the initial point at (0, 3).
Let's summarize what we've learned about solving explicit differential equations. These equations have the form dy/dx equals f of x and y, where the derivative is explicitly expressed. The solution method depends on the type of equation. For separable equations, we separate the variables and integrate both sides. For linear equations, we use the integrating factor method. Other types include homogeneous equations, where we substitute v equals y over x; exact equations, where we find a potential function; and Bernoulli equations, where we use the substitution v equals y to the power of 1 minus n. When solving Initial Value Problems, we first find the general solution containing an arbitrary constant C. Then we apply the initial condition to determine the specific value of C. Finally, we substitute this value back to obtain the particular solution that satisfies both the differential equation and the initial condition. These techniques form the foundation for solving more complex differential equations that arise in various fields of science and engineering.