On October 23rd, 2014, we updated our
By continuing to use LinkedIn’s SlideShare service, you agree to the revised terms, so please take a few minutes to review them.
Use initial conditions to find particular solutions of differential equations Use slope fields to approximate solutions of differential equations Use Euler’s method to approximate solutions of differential equations
General and Particular Solutions A function y = f(x) is called a solution of a differential equation if the equation is satisfied when y and its derivatives are replaced by f(x) and its derivatives. For example, is a solution to the differential equation y’ + 2y = 0. This can be shown by differentiating y. In general, is a general solution of y’ + 2y = 0. y’ + 2y = 0 is called a first-order differential equation. The order of a differential equation is determined by the highest-order derivative in the equation. In section 4.1, we showed that the second-order differential equation s”(t) = – 32 has the general solution s(t) = –16t 2 + C 1 t +C 2
Ex 1 p. 404 Verifying Solutions Determine whether the function is a solution of the differential equation Solution: Since y = cos x is not a solution. Solution: Since is a solution. Since is a solution. Solution:
Geometrically the general solution of a first-order differential equation represents a family of curves known as solution curves. For example, y = C/x is a solution to xy’ + y = 0 and we can see four solution curves with different C’s. Particular solutions are obtained from knowing some initial conditions that allow us to solve for unknown quantities.
Ex 2 p. 405 Finding a particular solution For the differential equation 3x+ 2yy’ = 0 and initial condition that y(1) = 3, verify the that the general solution 3x 2 +2y 2 = C satisfies the differential equation. Then find the particular solution that satisfies the initial condition. Solution: Find the derivative of 3x 2 +2y 2 = C With implicit differentiation, 6x + 4yy’ = 0. This factors to 2(3x +2yy’)= 0 so the differential equation is satisfied. Plugging in x=1 and y = 3, 3(1) 2 + 2(3) 2 = 3 + 18 = 21, so C = 21. The particular solution is 3x 2 +2y 2 = 21 Side note: the number of initial conditions must match the number of constants in the general solution.
Solving a differential equation analytically can be difficult or even impossible. There is a graphical approach you can use to find out a lot about the solution of a differential equation. Consider a differential equation in the form y’ = F(x,y) At each point (x,y) where F is defined, the differential equation determines the slope at that point. If you draw short segments representing this slope at several points, you can get an idea of what the solution curve looks like. These short segments form a slope field or a direction field for the differential equation y’ = F(x,y). This shows the general shape of all solutions. You can do these by hand by evaluating slope for every point, and sketching in a small segment representing this slope.
You can also do these by calculator: Download programs from me or list of program steps available on www.mrshartscalc.weebly.com under 2 nd trimester (not in powerpoints section, just 2 nd trimester top tab You need one called Slpfld and one called Euler a. b. c. Match each slope field with its differential equation: 1. y’ = x + y 2. y’ = x 3. y’ = y
From figure you can see that the slope at any point along the y-axis is 0. The only equation that fits this is #2
From the figure you can see that the slope at ( -1, 1) is 0. This fits #1.
From the figure you can see that at any point along the x-axis the slope is 0. This fits #3.
All these windows are from x: [-2,2] and y: [-2,2]
A solution curve of a differential equation y’ = F(x, y) is a curve in the xy-plane whose tangent line at each point (x, y) has slope equal to that of F(x, y) Ex 5 p. 407 Sketching a Solution Using a Slope-Field Sketch a slope field for the differential equation y’ = 2x + y Use the slope field to sketch the solution that goes through (1, 1) Solution: Make a table of values showing the slopes at several points. Then draw the line segments with their respective slopes at each point. To draw the particular solution going through (1, 1), start at (1, 1) and move to the right or left, allowing your curve to match slopes in near vicinity. x -2 -2 -1 -1 0 0 1 1 2 2 y -1 1 -1 1 -1 1 -1 1 -1 1 y’ = 2x+y -5 -3 -3 -1 -1 1 1 3 3 5