Differential Equations Agenda • Define terms • Learn how to separate a separable differential equation • Separate and solve 1st order separable differential equations. Differential Equations Definition Example A differential equation is an equation involving derivatives of an unknown function and possibly the function itself as well as the independent variable. y sin x , y ' 4 y 2 2 xy x 2 0, y y 3 x 0 1st order equations 2nd order equation The order of a differential equation is the highest order of the derivatives of the unknown function appearing in the equation In the simplest cases, equations may be solved by direct integration. Definition Examples y sin x y cos x C y 6 x e x y 3 x 2 e x C1 y x 3 e x C1x C2 Observe that the set of solutions to the above 1st order equation has 1 parameter, while the solutions to the above 2nd order equation depend on two parameters. Separable Differential Equations A separable differential equation can be expressed as the product of a function of x and a function of y. dy g x h y dx h y 0 Example: dy 2 xy 2 dx dy 2 x dx 2 y y 2 dy 2 x dx Multiply both sides by dx and divide both sides by y2 to separate the variables. (Assume y2 is never zero.) Separable Differential Equations A separable differential equation can be expressed as the product of a function of x and a function of y. dy g x h y dx Example: dy 2 xy 2 dx dy 2 x dx 2 y y 2 dy 2 x dx h y 0 2 y dy 2 x dx 1 y C1 x C2 2 1 x2 C y 1 2 y x C Combined constants of integration 1 y 2 x C Family of solutions (general solution) of a differential equation Example dy x dx y ydy xdx y 2 x2 C The picture on the right shows some solutions to the above differential equation. The straight lines y = x and y = -x are special solutions. A unique solution curve goes through any point of the plane different from the origin. The special solutions y = x and y = -x go both through the origin. Initial conditions • In many physical problems we need to find the particular solution that satisfies a condition of the form y(x0)=y0. This is called an initial condition, and the problem of finding a solution of the differential equation that satisfies the initial condition is called an initial-value problem. • Example (cont.): Find a solution to y2 = x2 + C satisfying the initial condition y(0) = 2. 22 = 0 2 + C C=4 y2 = x 2 + 4 Example: dy 2 x2 2 x 1 y e dx Separable differential equation 1 x2 dy 2 x e dx 2 1 y 1 u x2 x2 1 y 2 dy 2 x e dx du 2x dx 1 u dy e 1 y2 du tan 1 y C1 eu C2 1 tan y C1 e C2 x2 1 tan y e C x2 Combined constants of integration Example (cont.): dy 2 x2 2 x 1 y e dx 1 tan y e C x2 We now have y as an implicit function of x. tan tan y tan e C We can find y as an explicit function 1 y tan e C x2 x2 of x by taking the tangent of both sides. • Look at the equation and identify the x's and y's Treat dx as an x and dy as a y • Separate the the x's and y's,so that they are on opposite sides of the equation. • It is very important that one sided of the equation is multiplied by dx and the other by dy . In the end dx and dy should never be in the denominator. • Integrate both sides • Once we integrate we only need to add "+C" add to the left side of the equation • If possible, get our answer into y=f(x) form • Finally, if you are given an initial condition, solve for "+C" by plug in the given x and y value https://docs.google.com/forms/d/11 tMqcfAZ8TIGIMpfSQsYtqwLUUI etpgozHrhLn0e-M4/viewform