2.6 Not Exact - Integrating Factors

2.6 Not Exact - Integrating Factors - GE 207K Summary:...

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Unformatted text preview: GE 207K Summary: Exact Equations Integrating Factors September 13, 2011 A differential equation which can be written in the form M (x, y) + N (x, y) y = 0 . is not exact if My (x, y) = Nx (x, y) The following steps may be taken to find the general solution to the differential equation: Steps: 1. Write the given differential equation in the STANDARD FORM M (x, y) + N (x, y) y = 0 . 2. Is the integrating factor function of x or y? On exams, most likely you will be given this information. If you haven't been given this information, assume = (x), and if it did not work, try (y). Example 1 shows how having a wrong might not work. If is only a function of x, i.e., = (x), then, (x) = e R My -Nx N dx . (1) If is only a function of y, i.e., = (y), then, (y) = e R Nx -My M dx . (2) NOTE : Some professors won't give you credit for using above two equations, and require you to show where the equation for the integrating factor comes from. Consult your professor! If is not explicitly function of x or y, go back to the drawing board and figure out its form. 3. Multiply the standard form of the ODE by . Note that now your equation should be ? exact. (You can verify this by checking to see whether (M )y = (N )x ). 4. Proceed to find as you did in the case you had an exact equation. ! 1 c hf, 2011 GE 207K Exact Equations Integrating Factors September 13, 2011 Examples solved in class: Example 1 2xy dx + 4y + 3x2 dy = 0 Step 1 and 2: The standard form of the equation is dy 2xy + 4y + 3x2 = 0. dx M = 2xy N = 4y + 3x2 My = 2x Nx = 6x Therefore, My = Nx and therefore the equation is NOT exact. Step 3: Since we are not told whether is function x or x, let's assume = (x). M y - Nx 2x - 6x -4x = = 2 N 4y + 3x 4y + 3x2 = (x, y) = (x) Note that this is a function of x AND y. Therefore, we now have = (x, y) and our assumption that = (x) was incorrect. Let's assume = (y). Then we have Nx - M y 6x - 2x 2x 2 = = = . M 2xy 2xy y (y) = e Therefore, (y) = y 2 . Step 4: Multiply the ODE through by (y) = y 2 to get dy 2xy 3 + 4y 3 + 3y 2 x2 = 0. dx 2 R 2 y dy = e2lny = y 2 . Let's take a minute and verify that the above equation is now indeed exact. M = 2xy 3 N = 4y 3 + 3y 2 x2 My = 6xy 2 Nx = 6y 2 x c hf, 2011 GE 207K Exact Equations Integrating Factors September 13, 2011 My = Nx and therefore the equation is now exact. Step 5: Proceed to find . = M dx = 2xy 3 dx = x2 y 3 + h(y) (3) Step 6: Find h(y) by differentiating with respect to y and equating it to N : y = N y 2 + h (y) = 4y 3 + y 2 3x2 3x2 h(y) = y 4 Step 7: Recall that = C is the solution to the differential equation. = x2 y 3 + y 4 . x2 y 3 + y 4 = C . 3 c hf, 2011 GE 207K Example 2 Exact Equations Integrating Factors September 13, 2011 ydx + (x2 y - x)dy = 0, Step 1 and 2: The standard form of the equation is = (x) Therefore, dy y + x2 y - x = 0. dx M = y My = 1 N = x y - x Nx = 2xy - 1 2 My = Nx and therefore the equation is NOT exact. Step 3: In this problem, we are told that the integrating factor is a function of x, i.e. = (x). Using Eq. (1) (x) = e R My -Nx N dx =e R 1-2xy+1 x2 y-x dx =e R 2(1-xy) -x(1-xy) dx =e R -2 x dx = x-2 . Step 4: Multiply through by the integrating factor: 1 dy -2 =0 yx + y - x dx You can easily verify that the above equation is now exact. Step 5: Find . Not that both M and N are easy to integrate. So let's integrate M here: y x = M = M dx = yx-2 dx = - + h(y) x Note that we didn't forget to add the arbitrary function (not a constant)! Step 6: Find the arbitrary function h(y) by integrating previous expression with respect to y and equating it to N . y = N 1 1 - + h (y) = - + y x x y N h (y) = y h(y) = y2 2 c hf, 2011 4 GE 207K Exact Equations Integrating Factors September 13, 2011 Step 7: Recall that the solution to the given differential equation is = constant. y y2 - + =C. x 2 (4) 5 c hf, 2011 GE 207K Not Exact Equations Problem Set February 1, 2011 1. Solve the following first-order differential equation. y dx + x2 y - x dy = 0, = (x) Writing the equation in standard form we find dy y + x2 y - x =0 dx M = y , N = x2 y - x My = 1 , Nx = 2xy - 1 Comparing My and Nx , we note that My = Nx Not exact. (1) We are told that there exists an integrating factor which is a function of x. Therefore, M y - Nx 1 - 2xy + 1 2(1 - xy) 2 = = = - 2y - x N x -x xy) (1 - x (x) = e R My -Nx N dx = e- R 2 x dx = x-2 . Next, we multiply through by the integrating factor = x-2 to obtain. 1 dy -2 yx + y - = 0. x dx Make sure from now on you work with this equation! We can now either integrate M (x, y) with respect to x OR integrate N (x, y) with respect to y. Integrating M (x, y) with respect to x, we find, y (x, y) = M (x, y) dx = yx-2 dx = - + h(y) x Note that we didn't forget to include the arbitrary function (not a constant)! Recall that x = M and y = N . Therefore, differentiating previous equation with respect to y and equating it to N (x, y), we find 1 dh 1 - + =- +y dy x x x N ! ! y2 h= 2 1 c hf, 2011 GE 207K Not Exact Equations Problem Set February 1, 2011 And now we've found to its entirety! y y2 (x, y) = + x 2 Recall that the solution to an exact differential equation is (x, y) = C. Therefore, y y2 - + =C. x 2 (2) 2 c hf, 2011 GE 207K Not Exact Equations Problem Set February 1, 2011 2. Show that the following differential equation is not exact. Then find an integrating factor (x) and solve the differential equation. (xy - 1) dx + x2 - xy dy = 0. (3) As always, we begin by identifying the two functions M and N . Writing the equation in standard form we have dy (xy - 1) + x2 - xy = 0. dx M = xy - 1 , My = x , Comparing My and Nx , we note that My = Nx Not Exact. N = x2 - xy Nx = 2x - y And we note that The problem statement tells us that there exists an integrating factor such that it is only a function of x. Therefore, M y - Nx x - 2x + y (-x 1 + y) = = = -x N x2 - xy -x y) (-x + R My -Nx R 1 1 dx N (x) = e = e- x dx = x-1 = . x So we've found the integrating factor. Don't get carried away just yet! We've solved about half the problem so far. Multiply the ODE by the integrating factor to make it exact: 1 dy y- + (x - y) = 0. x dx We can double-check our work to see if we've found the right integrating factor. Of course, ? we'd do this by checking whether My = Nx , where certainly M and N are our new functions. Now that we have an exact equation, we proceed to go on about our work and find . We can either integrate M (x, y) with respect to x OR integrate N (x, y) with respect to y. Integrating M (x, y) with respect to x, we find, 1 (x, y) = M (x, y) dx = y - dx = yx - ln |x| + h(y) x 3 c hf, 2011 GE 207K Not Exact Equations Problem Set February 1, 2011 Recall that x = M and y = N . Therefore, differentiating previous equation with respect to y and equating it to N (x, y), we find x + h (y) x h = -y Therefore, = - y x N h=- y2 2 (x, y) = yx - ln x - y2 . 2 Recall that the solution to an exact differential equation is (x, y) = C. Therefore, yx - ln x - y2 =C. 2 (4) 4 c hf, 2011 GE 207K Not Exact Equations Problem Set February 1, 2011 3. Show that the following differential equation is not exact. Then find an integrating factor (y) and solve the differential equation. 2y dx + (x + y) dy = 0. (5) As always, we begin by identifying the two functions M and N . Writing the equation in standard form we have dy (2y) + (x + y) = 0. dx And we note that M = 2y , N = x + y My = 2 , Nx = 1 Comparing My and Nx , we note that My = Nx Not Exact. The problem statement tells us that there exists an integrating factor such that it is only a function of y. Therefore, Nx - M y 1 =- M y 1 1 = e- 2 ln |y| = . y (y) = e R Nx -My M dy = e- R 1 y dx Multiply the ODE by the integrating factor to make it exact: x dy 2 y+ + y = 0. y dx M N It can easily be verified that the new equation is now exact. Now that we have an exact equation, we proceed to go on about our work and find . We can either integrate M (x, y) with respect to x OR integrate N (x, y) with respect to y. Integrating M (x, y) with respect to x, we find, (x, y) = M (x, y) dx = 2 y dx = 2x y + h(y) Recall that x = M and y = N . Therefore, differentiating previous equation with respect to y and equating it to N (x, y), we find x x + h (y) = + y y y x N h = y 2 2 h = y3 3 5 c hf, 2011 GE 207K Therefore, Not Exact Equations Problem Set February 1, 2011 2 2 (x, y) = 2x y + y 3 . 3 Recall that the solution to an exact differential equation is (x, y) = C. Therefore, 2 2 2x y + y 3 = C . 3 (6) 6 c hf, 2011 GE 207K Not Exact Equations Problem Set February 1, 2011 4. Show that the following differential equation is not exact. Then, find an integrating factor (y) that will turn the differential equation exact. Finally solve the differential equation. 2 y + 2xy dx - x2 dy = 0 (7) As always, we begin by identifying the two functions M and N . Writing the equation in standard form we have 2 dy y + 2xy + -x2 = 0. dx And we note that M = y 2 + 2xy , N = -x2 My = 2y + 2x , Nx = -2x Comparing My and Nx , we note that My = Nx Not Exact. The problem statement tells us that there exists an integrating factor such that it is only a function of y. Therefore, Nx - M y -2x - 2y - 2x -2 (2x + y) 2 = = =- 2 + 2xy M y y 2x) (y + y R 1 R Nx -My 1 dy M (y) = e = e-2 y dx = 2 . y Multiply the ODE by the integrating factor to make it exact: 2 2x x dy 1+ + - 2+ = 0. y y dx M N It can easily be verified that the new equation is now exact. In fact, let's do it: M =1+ 2x x2 , N =- 2 y y 2x 2x My = - 2 , Nx = - 2 y y M y = Nx Exact. Now that we have an exact equation, we proceed to go on about our work and find . We can either integrate M (x, y) with respect to x OR integrate N (x, y) with respect to y. Integrating M (x, y) with respect to x, we find, 2x x2 (x, y) = M (x, y) dx = 1 + dx = x + + h(y) y y 7 c hf, 2011 GE 207K Not Exact Equations Problem Set February 1, 2011 Recall that x = M and y = N . Therefore, differentiating previous equation with respect to y and equating it to N (x, y), we find 2 2 x x - 2 + h (y) = - 2 y y x N h =0 Therefore, h = c1 (x, y) = x + x2 + c1 . y Recall that the solution to an exact differential equation is (x, y) = C. Therefore, x+ x2 =C. y (8) 8 c hf, 2011 ...
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This note was uploaded on 02/03/2012 for the course M 427K taught by Professor Fonken during the Fall '08 term at University of Texas at Austin.

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