Solutions of separable equations often are in an implicit form; that is, not solved for the dependent variable.

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1 Section 2.2 Separable DEs Key Terms: Separable DEs; the DE can be rearranged so that all terms containing the independent variable are on one side of the equal sign and all terms containing the dependent variable are on the other side. Solutions of separable equations often are in an implicit form; that is, not solved for the dependent variable. Sometimes we cannot get an explicit expression for the solution.

2 A first order ODE is given by y'(x) = f(x,y) If the right side f(x, y) can be written as a product of two functions, one depending upon x alone and the other depending upon y alone, then we say the DE is separable. That is, if Sometimes it is convenient to express h(y)dy = g(x)dx in the form h(y)dy g(x)dx = 0, and integrate each term and the right side. Previously we used t for the independent variable, many times first order DEs are in terms of x and y.

3 Examples: Obtain a separated expression for the DE.

4 Example: Solve the IVP, give an explicit solution (if possible), show some integral curves and discuss the domain of the solution. Separate the variables to get Integrate both sides to get Apply the initial condition: set x = 0 and y = -1 We get 3 = C. We have implicit solution In this case we can solve for y using the quadratic formula to get an explicit expression. To satisfy the initial condition we must choose the sign because y(0) = -1.

5 Thus our explicit solution is To determine the domain of our solution we require x s so that Use some algebra: 3 2 x 2x 2x 4 2 x (x 2) 2(x 2) 2 (x 2)(x 2) This is greater than or equal to zero only when x+ 2 0; so x x 3 +2 x 2 +2 x+4 From the graph we see that x -2. (It can be shown that this is the only root (or zero) of the cubic polynomial.) x

6 1 0.5 Solution of IVP 1-sqrt(x. 3 +2*x. 2 +2*x+4) Y-axis Explicit solution: x in [-2, 2]. What is going on at x = -2? X-axis

7 The graph of the integral curve corresponding to the solution of the IVP is 3 Implicit plot of y 2-2*y-(x 3 +2*x 2 +2*x+3)=0 2 Vertical tangent at y = (0, -1) Graph done using MATLAB.

8 Some integral curves for the DE are shown. 2 dy 3x 4x 2 dx 2(y 1) What happens when y = 1? We have a vertical tangent at x = -2. This restricts the domain since the derivative is undefined. IVP curve Initial condition Copy from the text book.

9 Example: Solve the IVP, give an explicit solution (if possible), show some integral curves and discuss the domain of the solution. Separate the variables to get Integrate both sides to get Multiply both sides by 4 and rearrange to get Apply the initial condition: set x = 0 and y = 1 to get C = 17. Thus an implicit solution is Here there is no reasonable way to solve for an explicit form for the solution.

10 The direction field shows the solution of the IVP is valid on either side of the initial point until the slope becomes undefined (vertical). We see that the interval ends when we reach points where the tangent line is vertical. It follows from the differential equation that these are points where 4 + y 3 = 0. Let s investigate this further so we can find the domain of the solution.

11 The tangent line is vertical. where 4 + y 3 = 0. So that is where y = (-4) 1/ Next we find the corresponding values of x using the solution Substituting the value for y into the solution we get The roots of this polynomial are We are only interested in the real roots so the domain of the solution is approximately [ , ].

12 Other integral curves are shown

13 Application: The rate at which a mothball evaporates from solid to gaseous state is proportional to the surface area of the mothball. Suppose at time t = 0 the mothball has a radius of 0.5 inch and after 6 months a radius of 0.25 inch. Construct a differential equation to describe the rate of change of the mothball as it evaporates and determine the proportionality constant. Finally predict the size of the mothball after 1 year. Mothballs are small balls of chemical pesticide and deodorant sometimes used when storing clothing and other articles susceptible to damage from mold or moth larvae (especially clothes moths). Ref: Write an IVP for this situation: Let r = radius of the mothball dr/dt = k(surface area of a sphere), r(0) = 0.5 in. Ref: Farlow dr = k(4 πr 2 ), r(0) = 0.5 dt

14 Separate the variables: dr 2 = k(4 π)dt r Integrate both sides: Apply initial condition r(0) = 0.5 to find C. C = = 4k π t + C r 1 - = 4k π t - 2 r 1 - = 4k π t - 2 r Use r(6) = 0.25 to determine k; set t = 6 and r = = 4κ π 6-2 ==> 2-4 = 4kπ 0.25 so -2 k= 24 π = 4 π t - 2 = t - 2 r 24 π 3 Now set t = 12 months and solve for r: when t = 12 months r = 1/6 inch.