Separable Variables, differential equations, and graphs of their solutions This will be an eploration of a variety of problems that occur when stuing rates of change. Many of these problems can be modeled by first order differential equations with separable variables. These equations can be written in the form = f ( ) g(y ). These need to be rearranged as g(y ) that y = y( ). Then, g(y( )) = f ( ), recalling the Chain Rule and the fact = f ( ) and g(y ) = f ( ) + c. EXAMPLE: Solve the following separable differential equation = y = y = y y = = y + = c A Slope- Field demonstrates the gradient to the curve described by y = y( ).
EXAMPLE: Show that the general solution to the differential equation y = can be written as y= + Ae if y>. y = = y ln y = + c y = e + c + c since y > 0, y = e = e e y = + Ae y π EXAMPLE: Solve the differential equation cos = e for 0 < given that π =, y= giving your answer in the form () 4 y= f. = sec y e y e = sec y e = sec y y e = tan+ c e = tan+ c e= + c c= e y e = tan+ e y= ln tan+ e y= ln tan+ e c
Modelling Growth and Decay phenomena An important class of separable differential equation has the form =± ky where dt k > 0 and t is the independent variable, time. =± ky =± kdt dt y ln y =± kt + c kt c kt y = e ± + y = Ae ± The general solution y= Ae ±kt is called the eponential growth curve. This model is used to calculate continuous compound interest as well as to represent the eponential decay curve, such as in radio carbon dating, used to measure age by determining the amount of C 4 remaining. The half- life of carbon is 5568 years. EXAMPLE: The Lascau Caves in southwest France contain some of the oldest and finest prehistoric art in the world. By means of comple chemical analysis on charcoal taken from the caves, scientists were able to determine that the charcoal contained 5% of the original amount of C 4 that it would have contained when the tree it was made from was cut down. a) Find the value of the decay constant k for the model that gives the quantity Q of C 4 present in the sample of charcoal. b) Hence, find an approimate age for the Lascau paintings. a) kt Qt () = Q e 0 5668k Q( 5568) = Q0e = Q0 ln 5668k= ln k= 0. 0004 5668 b) kt ln( 05. ) Qe 0 = 05. Q0 t= k t 500 years
TRY THIS Artjom has been eploring economic models for inflation. He decides to consider a model where the inflation decreases over time according to the dp I0 model = Pwhere t is the time in years, P(t) is the price of an item at a dt I + t time t and I 0 is the initial inflation rate. a) Assuming that the inflation rate starts at %, calculate the price of an item in 0 years if P(0)=00. b) Find an epression for P(0) in terms of I 0 given that P(0)=00. John thinks that actually the inflation will increase in the near dp future and suggests a different model: (.. t) P dt = 00 + 000 c) Solve the differential equation and find the value of P(0) given that P(0)=00. FIRST ORDER EXACT EQUATIONS AND INTEGRATING FACTORS The equation y + = is an eample of an eact equation. It is a linear differential equation as it can be written as y + =. You may also notice that d d + y= ( y) which means that the equation can be written as ( y ) =. c Integrating with respect to : ( y) = + c y =, X + 0. d Equations that can be written in the form ( u ( ) y ) = v ( ) are called first order eact equation. Most first order equations are not eact, but can be transformed by multiplying both sides of the equation by an appropriate epression called an integrating factor.
For eample, y y + = is not eact but if you multiply both sides by you get y + y = 6 d ( y ) = 6 or d ( y ) = 6 which is eact. If we can find an integrating + py () = q () d p( ) p( ) e y = e q () factor that makes a first order equation eact, then the problem of solving it becomes simply an integration problem. Fortunately, for first order linear equations of the form + py () = q (), we can find the integrating factor in a systematic way as stated in the following theorem: Theorem: Given the differential equation + py () = q (), the function ( ) I () e p = is an integrating factor that transforms this differential equation into and d p ( ) p ( ) eact differential equation of the form e y = e q () Proof: Multiply both sides by I(): p ( ) p ( ) p ( ) e + p() e y= e q() d d e = p() e p() e = p ( ) p ( ) d p ( ) e + p() e y= e y d p ( ) p ( ) e y = e q () p ( ) p ( ) p ( ) d d e = p() e p() e = p( ) p( ) d p( ) e + p() e y= e y p( ) p( ) p( ) d p( ) p( ) e y = e q ()
EXAMPLE: Consider the first order linear equation y e + = a) Find an integrating factor for this differential equation. b) Hence, solve the differential equation. a) b) EXAMPLE: y e + y= e + =, 0 ln I () e = = e = y e + = d ( y ) e = y= e + c c y= e +, 0 Find the particular solution of the following first order differential equation that satisfies the initial condition, given y e and y( ) + = 0 = e + y = e e e + e y= e d e y e e y e = = + c c y= e + = e + ce e