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Unformatted text preview: GE 207K Exact Equations September 8, 2011 Summary: A differential equation which can be written in the form M (x, y) + N (x, y) y = 0 . is exact IF My (x, y) = Nx (x, y) Once a differential equation has been identified as exact, then there exists a function such that it satisfies the following two conditions: x = M (x, y) , y = N (x, y) . The following steps can 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. Check for exactness: If My = Nx exact. If not, not exact. (1) 3. Recall that x = M and y = N . Either integrate M with respect to x OR integrate N with respect to y to find . Choose which ever is easier to integrate. So: = M dx + h (y) (2) OR = N dy + h (x) (3) Remember to add an arbitrary function (not a constant) after integrating as shown above. 4. To find the arbitary function h, differentiate you obtained and equating it with the corresponding function M or N : If you chose = M dx + h (y) then differentiate with respect to y and equate the resulting expression to N . Find what h (y) is by comparing the two sides. If you chose = N dy + h (x) then differentiate with respect to x and equate the resulting expression to M . Find what h (x) is by comparing the two sides. 5. Solution is =C. 1 (4) c hf, 2011 ! GE 207K Examples solved in class: Example 1 Exact Equations September 8, 2011 (1  y sin x) dx + (cos x) dy = 0 Step 1: Find functions M and N and check for exactness. M = 1  y sin x N = cos x My =  sin x Nx =  sin x My = Nx and therefore the equation is exact. Step 2: Find function . Note that we can integrate either M or N with ease. So let's choose to integrate M with respect to x: x = M = M dx = (1  y sin x dx) dx = x + y cos x + h (y) where h (y) is a result of integrating the multivariable function with respect to x. Step 3: Differentiate Eq. (5) with respect to y and recall that y = N cos x + h (y) y (5) Step 4: Determine h (y) by inspecting Eq. (6). = cos x N (6) h (y) = 0 h (y) = c1 Step 5: Recall that = C is the solution we seek. = x + y cos x + c1 (7) x + y cos x = C . Note that c1 is taken into account by the constant C. 2 c hf, 2011 GE 207K Example 2 Exact Equations September 8, 2011 x3 + 3xy 2 dx + 3x2 y + y 3 dy = 0 M = x3 + 3xy 2 N = 3x2 y + y 3 My = 6xy Nx = 6xy Step 1: Find functions M and N by inspection, and check for exactness: My = Nx and therefore the equation is exact. Step 2: Find function . Note that we can integrate either M or N with ease. So let's choose to integrate M with respect to x: x = M = M dx 3 = x + 3xy 2 dx dx = x4 3 2 2 + x y + h (y) 4 2 where h (y) is a result of integrating the multivariable function with respect to x. Step 3: Differentiate Eq. (8) with respect to y and recall that y = N 3x2 + y h (y) y (8) Step 4: Determine h (y) by inspecting Eq. (9). = y + y 3 3x2 N (9) h (y) = y 3 y4 h (y) = + c1 4 Step 5: Recall that = C is the solution we seek. = x4 3 2 2 y 4 + xy + + c1 4 2 4 (10) x4 3 2 2 y 4 + xy + =C . 4 2 4 Note that c1 is taken into account by the constant C. 3 c hf, 2011 GE 207K Exact Equations Problem Set September 8, 2011 Solve the following firstorder differential equations. 1. (1  y sin x) dx + cos x dy = 0 Writing the equation in standard form we find (1  y sin x) + cos x dy = 0, dx M = 1  y sin x , N = cos x My =  sin x , Nx =  sin x Comparing My and Nx , we note that M y = Nx Exact. 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 = (1  y sin x) dx = x + y cos x + 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 cos x + Therefore, (x, y) = x + y cos x + c1 Recall that the solution to an exact differential equation is (x, y) = C. Therefore, x + y cos x + c1 = C x + y cos x = c2 , where we combined constants c1 and C into c2 .
x dh = x cos dy N h = c1 (1) 1 c hf, 2011 GE 207K Exact Equations Problem Set September 8, 2011 2. Is the following differential equation exact? If so, solve it. y x dx + dy = 0 x2 + y 2 x2 + y 2 Writing the equation in standard form we have x y dy + =0 x2 + y 2 x2 + y 2 dx x y M= , N= x2 + y 2 x2 + y 2 xy xy My =  2 , Nx =  2 2 )3/2 (x + y (x + y 2 )3/2 Comparing My and Nx , we note that M y = Nx Exact. (2) And we note that 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 (x, y) = M (x, y) dx = dx = x2 + y 2 + h(y) x2 + y 2 1/2 1 2 y x 2 (2y) + h (y) = + y 2 x2 + 2 y x Recall that x = M and y = N . Therefore, differentiating previous equation with respect to y and equating it to N (x, y), we find N h =0 h=c Therefore, (x, y) = x2 + y 2 + c Recall that the solution to an exact differential equation is (x, y) = C. Therefore, x2 + y 2 + c = C x2 + y 2 = c2 , where we combined constants c1 and C into c2 . 2 (3) c hf, 2011 GE 207K Exact Equations Problem Set September 8, 2011 3. Solve the following differential equation. u2 2u 2v du + 2 dv = 0 2 +v u + v2 (4) The standard form of the differential equation is u2 And we note that M= Mv =  (u2 u2 2u + v2 , 2v + v2 4uv Nu =  2 (u + v 2 )2 , N= u2 2u 2v dv + 2 = 0. 2 +v u + v 2 du 4uv + v 2 )2 Comparing Mv and Nu , we note that M v = Nu Exact. We can either integrate M (u, v) with respect to u OR integrate N (u, v) with respect to v. Integrating M (u, v) with respect to u, we find, 2u (u, v) = M (u, v) du = du = ln u2 + v 2  + h(v) 2 + v2 u Recall that u = M and v = N . Therefore, differentiating previous equation with respect to v and equating it to N (u, v), we find 2v v 2 + h (v) u2 + v = h =0 h=c 2v v2 u2 +
N Therefore, (u, v) = ln u2 + v 2  + c Recall that the solution to an exact differential equation is (u, v) = C. Therefore, ln u2 + v 2  + c = C where we combined constants c1 and C into c2 . 3 c hf, 2011 u2 + v 2 = c 2 , (5) GE 207K Exact Equations Problem Set September 8, 2011 NOTE: Alternatively, all this work could have been avoided if we had realized that we can multiply the ODE through by u2 + v 2 to obtain 2udu = 2vdv, which is a separable equation. We then would find u2 + v 2 = C, which is the same answer as before. (6) 4 c hf, 2011 GE 207K Exact Equations Problem Set September 8, 2011 4. Solve the following differential equation.
y 2 u dy = du ln u (7) The standard form of the differential equation is y  2 du + (ln u) dy = 0. u And we note that M = 2 + y , N = ln u u 1 1 My = , Nu = u u Comparing My and Nu , we note that M y = Nu Exact. We can either integrate M (y, u) with respect to u OR integrate N (y, u) with respect to y. Integrating M (y, u) with respect to u, we find, y (y, u) = M (y, u) du = 2 + du = 2u + y ln u + h(y) u Recall that u = M and y = N . Therefore, differentiating previous equation with respect to y and equating it to N (y, u), we find ln u + h (y) h =0 Therefore, y = ln u N h=c (y, u) = 2u + y ln u + c Recall that the solution to an exact differential equation is (y, u) = C. Therefore,  2u + y ln u + c = C 2u + u ln u = c2 , where we combined constants c1 and C into c2 . (8) 5 c hf, 2011 GE 207K Exact Equations Problem Set September 8, 2011 5. Solve the following differential equation. (sin 2t) dx + (2x cos 2t  2t) dt = 0 The standard form of the differential equation is (2x cos 2t  2t) + sin 2t And we note that M = 2x cos 2t  2t , N = sin 2t Mx = 2 cos 2t , Nt = 2 cos 2t Comparing Mx and Nt , we note that M x = Nt Exact. dx = 0. dt (9) We can either integrate M (x, t) with respect to t OR integrate N (x, t) with respect to x. Integrating M (x, t) with respect to t, we find, (x, t) = M (x, t) dt = (2x cos 2t  2t) dt = x sin 2t  t2 + h(x) Recall that t = M and x = N . Therefore, differentiating previous equation with respect to x and equating it to N (x, t), we find + h (x) sin 2t h =0 Therefore, x h=c = 2t sin N (x, t) = x sin 2t  t2 + c Recall that the solution to an exact differential equation is (x, t) = C. Therefore, x sin 2t  t2 = c2 , where we combined constants c1 and C into c2 . x sin 2t  t2 + c = C (10) 6 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.
 Fall '08
 Fonken
 Differential Equations, Equations

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