MEC702_Wk_3_Lect_3 - MEC702 Wk 3 Lect 3 Separable ODEs...

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Unformatted text preview: MEC702 Wk 3 Lect 3 Separable ODEs. Modeling. Separable ODEs Many ODEs can be reduced to the form g(y)y 0 = f (x) by purely algebraic manipulations. Then integrate both sides with respect to x to obtain ˆ ˆ g(y)dy = g(x)dx + c . This method of solving ODEs is called method of separating variables and the equation above is called a separable equation, because the variables are now separated. Example. The ODE y 0 = 1 + y 2 is separable because dy dy = 1 + y 2 =⇒ = dx. dx 1 + y2 Hence, ˆ dy 1 + y2 tan−1 y y ˆ = dx = x+c = 1 tan(x + c). Example. Mixing Problem (Modeling) Tank contains 1000 gal of water in which initially 100 lb of salt is dissolved. Brine runs in at a rate of 10 gal/min and each gallon contains 5 lbs of dissolved salt. Mixture in the tank is kept uniform by stirring. Brine runs out at 10 gal/min. Find the amount of salt in the tank at any time t. Solution. Step 1: Setting up a model. Let y(t) = amount of salt in the tank at time t. Then the time rate of change is y 0 = Salt inow rate − Salt outow rate from Balance Law. So inow: outow: Our model is the ODE Step 2: Solution of the model. ODE is separable, so 5lb × 10gal = 50lb of salt. 10 = 0.01 or 1% of 10gal brine which is 1000 total brine content in the tank, i.e., 0.01y(t). y 0 = 50 − 0.0y = −0.01(y − 5000). ˆ ˆ dy = −0.01dt y − 5000 ln |y − 5000| = −0.01t + c∗ y 5000 + Ae−0.01t . = Initially, the tank contains 100lb of salt so y(0) = 100 is the initial condition. Thus, 100 = 5000 + ce0 =⇒ c = −4900. Hence, the amount of salt in the tank at time t is y(t) = 5000 − 4900e−0.01t . 2 Reduction to Separable Form Some ODEs are nonseparable but with careful transformation (i.e., substitution) they become separable. We look at the class of ODEs of the form y0 = f y x , where f is any dierential function of xy . Set u = xy . Then y = ux y 0 = u0 x + u . =⇒ Upon substitution, we have y0 = u0 x + u = du x = dx 1 du = f (u) − u f y x f (u) f (u) − u 1 dx x which is separable!!! Example. Given 2xyy 0 = y0 = y0 = y 2 − x2 y 2 − x2 2xy   1 y 1 x − 2 x 2 y Now substitute y = ux and y 0 = u0 x + u and then simplify to get u0 x + u = u0 x Then = u 1 − 2 2u −u2 − 1 . 2u ˆ ˆ 2u 1 du = − dx 1 + u2 x 1  ln 1 + u2 = − ln |x| + c∗ = ln + c∗ x c 2 1+u = x  y 2 c 1+ = x x x2 + y 2 = cx    c 2 c 2 x− + y2 = 2 2 a family of circles passing through the origin centering at the x-axis. 3 (1) Exact ODE's. Integrating Factors. Dierential Note that for a function y = f (x), its derivative dy = f 0 (x) dx is its total derivative. What about for functions z = f (x, y)? We have partial derivatives fx (x, y) = ∂f ∂x fy (x, y) = ∂f . ∂y and And its total derivative is dened using both of the partial derivatives df = ∂f ∂f dx + dy ∂x ∂y (2) It follows from this that if f (x, y) = c a constant, then du = 0. Example. If then f = x + x2 y 3 = c,   du = 1 + 2xy 3 dx + 3x2 y 2 dy = 0 or y0 = dy 1 + 2xy 3 =− , dx 3x2 y 2 an ODE we can solve by going backwards!!! Exactness A rst-order ODE M (x, y) + N (x, y)y 0 = 0 is also written as M (x, y)dx + N (x, y)dy = 0 is also called an exact dierential equation if its left side M (x, y)dx + N (x, y)dy is exact, i.e., it forms exactly the total dierential df = ∂f ∂f dx + dy ∂x ∂y of some function f (x, y). Hence Equation (3) can be rewritten as df = 0. Then by integration, we obtain a general solution of Equation (3) in the form f (x, y) = c. 4 (3) Implicit Solution and Contours f (x, y) = c This is called an implicit solution. (Note that the solution of the form y = h(x) is an explicit solution.) The graphs of f (x, y) = c are contour lines. Derivation of the Exactness Condition Comparing (2) and (3), we see that the ODE is exact when M= ∂f ∂x N= ∂f , ∂y but we do not know the function f (x, y) to use these conditions. (Equality of Partial Derivatives) Let f (x, y) be a function with continuous partial derivaand ∂f ∂y . Then Theorem. tives ∂f ∂x ∂ ∂x  ∂f ∂y  = ∂ ∂y  ∂f ∂x  Using the theorem from Multivariable Calculus given above, we obtain ∂N ∂M = ∂x ∂y which we will use from here onwards as the condition for an exact ODE. Finding f (x, y) for Exact ODE There are two ways of nding f (x, y) for an Exact ODE: 1. Integrating M = ∂f ∂x with respect to x yields ˆ f (x, y) = M dx + k(y) where k(y) is a hidden function in terms of y which we haven't found yet. To obtain k(y), we nd ∂f ∂y and equate ∂f ∂y N= to nd k0 (y). The integrating k0 (y), we nd the hidden function k(y). 2. Integrating N = ∂f ∂y with respect to y yields ˆ f (x, y) = N dy + h(x) where h(x) is a hidden function in terms of x which we haven't found yet. To obtain h(x), we nd ∂f ∂x and equate M= ∂f ∂x to nd h0 (x). The integrating h0 (x), we nd the hidden function h(x). 5 (4) Example. Consider 0 = cos(x + y)dx + (3y 2 + 2y + cos(x + y))dy. Step 1: Test for Exactness. We notice from Equation (3) that M = cos(x + y) N = 3y 2 + 2y + cos(x + y). Then checking for exactness (4): ∂M ∂y ∂N ∂x = − sin(x + y) = − sin(x + y). Hence ODE is exact. Step 2: Implicit General Solution. By integration, ˆ f = M dx + k(y) ˆ = = cos(x + y)dx + k(y) sin(x + y) + k(y). By partial dierentiation with respect to y and knowing that it is equal to N . N = ∂f dk = cos(x + y) + ∂y dy Hence, dk dy y Therefore, = 3y 2 + 2y = y 3 + y 2 + c∗ . f (x, y) = sin(x + y) + y 3 + y 2 = c. Step 3: Checking an Implicit Solution. By dierentiation, we check f (x, y) = c implicitly to see whether it leads to the given ODE. df = 0 = ∂f ∂f dx + dy ∂x ∂y   [cos(x + y)] dx + (3y 2 + 2y + cos(x + y)) dy. 6...
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