Section 2.2 Solutions to Separable Equations

Transcription

1 Section. Solutions to Separable Equations Key Terms: Separable DE Eponential Equation General Solution Half-life Newton s Law of Cooling Implicit Solution (The epression has independent and dependent variables combined; function of two variables.) Eplicit Solution (Dependent variable is epressed as function of the independent variable alone.)

2 Definition: A first order DE is called separable if it 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. One equation that arises often in applications is N' = -λn Because of the form of its solutions, the DE is called the eponential equation. An equivalent epression is dn =-λn dt. Separate the DE into the form dn =-λdt N Integrating both sides we get ln N = λt - + C -λt+c C -λt Solving for N gives N =e =e e are both positive we have two cases C To simplify things let e, if N > 0 A= C -e, if N < 0. Since e C and e -λt C λt - e e, if N > 0 N(t) = C λt - -e e, if N < 0 Thus the solution can be written N(t) = Ae -λt where A is a nonzero arbitrary constant. This procedure is not valid when N = 0, but note that if we set A to zero then N(t)= 0 and we have a solution for all values of A.

3 Eample: Radioactive decay The eponential equation N(t) = Ae -λt is a model for the decay of Radioactive Thorium- 34. Suppose that 1 gram of this material is reduced to 0.80 gram in one week. Find the half-life of T-34. Let N(t) be the amount of Thorium-34 at time = t weeks. We have N(0) = 1 gm and N(1) = 0.80 gm. There are two unknowns in the eponential equation: A and λ. Using N(t) = Ae -λt and N(0) = 1 we get 1 = Ae -λ0 = A. Thus A = 1. Using N(t) = e -λt and N(1) = 0.80 we get 0.80 = e -λ1 and solving for λ we have -λ = ln(0.80) λ = ln(0.80) Then we have solution N(t) = e ln(0.80)t. To find the half-life we set N = 0.50 gm and solve for t. ln(0.80) t ln(0.50) 0.50 = e ln(0.50) = ln(0.80)t t = weeks ln(0.80) So the half-life is about 3.1 weeks.

4 Eample: Solve DE y'= t y. Let y' = dy/dt and separate the variables. dy = t dt y 1 1 Integrating both sides we get - = t +C y Solving for y we obtain y= = = 1 t +C t +C t +C where we used that times an arbitrary constant is just an arbitrary constant which we renamed to be C. Note that the constant function y(t) = 0 is another solution although no finite value of C in the solution equation will yield this solution. Definition: The general solution to a differential equation is a family of solutions depending on sufficiently many parameters to give all but finitely many solutions. - y(t) = t +C So is general solution of y'= t y, even though this family of solutions doesn t include all solutions.

5 Look at the variety of solutions to IVPs for DE y'= t y. This is just a portion of the direction field. y = -/(t +C) C = -5, -6, y = 0 0 C = C = 1 C = 1,,

6 y ' = t y y t

7 Eample: A can of Coors Light beer at 40 F is placed into a room where the temperature is 70 F. After 10 minutes the temperature of the beer is 60 F. What is the temperature of the beer as a function of time? What is the temperature of the beer 0 minutes after the beer was placed into the room? According to Newton s law of cooling, the rate of change of an object s temperature (T ) is proportional to the difference between its temperature and the ambient temperature (A). Thus we have dt = -k(t - A), k > 0 dt Notice that if T < A, the temperature of the object will be increasing. The equation is separable, so we separate variables to get Integrate both sides to get dt (T - A) ln( T - A ) = -kdt = -kt + C We solve for T using eponentials to obtain -kt +C C -kt -kt -kt T - A = e = e e = Ce T = Ce + A After the integration step we were able to get an eplicit epression for T.

8 -kt T = Ce + A Using the information that at t = 0, T = 40 F and A = 70 F we determine C. -k0 40 = Ce + 70 C = -30 Now we have -kt T = -30e + 70 Using that after 10 minutes the temperature of the beer is 60 F we solve for k. -k10 -k10 1 -k10 ln(1/ 3) 60 = -30e = -30e = e k = Finally to determine the temperature of the beer 0 minutes after the beer was placed into the room we determine T from t T = -30e + 70 We find that ( 0) T = -30e So the beer is about 67 F after 0 minutes, which is almost room temperature.

9 Implicitly defined solutions It is not always possible to get an eplicit epression for the solution after you solve the separable DE. We consider eamples. Eample: Find the solutions of the equation y = e /(1+y), having initial conditions y(0) = 1 and y(0) = 4. When we separate the variables and integrate we obtain Implicit solution. 1 y+ y =e +C y +y-(e +C)=0 Applying the quadratic formula we get 1 y() = - ± 4 + 8(e + C) = -1± 1+ (e + C) We get two solutions from the quadratic formula, and the initial condition will dictate which solution we choose. Case: y(0) = 1 0 1= -1± 1+ (e + C) C = 1/ Inspection says we must choose the positive root so we have solution. Case: y(0) = = -1± 1+ (e + C) C = 3 Inspection says we must choose the negative root so we have solution. y() = (e +1/ ) = e y() = (e + 3) = e

10 We have the following sketch of the solutions. What is the interval of eistence for each solution? y() = e y() = e In each case the term under the square root is positive for all hence both solutions have interval of eistence (-, ).

11 In the previous eample we were able to take the implicit solution and use the quadratic formula to get eplicit solutions for each initial condition. This not always the case. Eample: Find the solutions of the equation (0) = 1, (0) =, and (0) = 0. t ' = 1+ having initial conditions When we separate the variables and integrate we obtain ( ) +ln = t + C It is not possible to obtain an eplicit epression in this case. Case: (0) = 1 C = 1 so we have ( ) +ln = t +1 Case: (0) = - C = ln()- so we have ( ) +ln = t +ln() - Case: (0) = 0 We can t separate the variables; we would have (1+ )d = tdt Note the denominator. However inspecting the IVP t ' =, (0) = 0 1+ we see that (t) = 0 is a solution.

12 Using numerical methods we can obtain a plot like the following. Since the solutions are defined implicitly, it is a difficult task to visualize them without the aid of numerical methods.