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13. Assume that equation (6) meets the requirements of Theorem 2.6.1 in a rectangle R and is therefore exact. Show that a possible function(x, y) is y X (x, y) = - [NC M(s, yo) ds + yo XO where (xo, yo) is a point in R. N(x, t) dt,

13. Assume that equation (6) meets the requirements of Theorem 2.6.1 in a rectangle R and is therefore exact. Show that a possible function(x, y) is y X (x, y) = - [NC M(s, yo) ds + yo XO where (xo, yo) is a point in R. N(x, t) dt,

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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4. dy dx ax + by bx + cy - by bx - cy ax- +(6y²-x²+3) y' = 0 5. 6. (yexy cos(2x)-2exy sin(2x)+2x)+(xexy cos(2x) - 3) y' = 0 7. (y/x+6x) + (lnx-2) y' = 0, x > 0 X y dy (x² + y²) 3/2 dx = 0 + 8. (x² + y2)3/2 In each of Problems 9 and 10, solve the given initial value problem and determine at least approximately where the solution is valid. 9. (2x - y) + (2y - x) y' = 0, y(1) = 3 10. (9x² + y − 1) − (4y − x) y' = 0,_y(1) = 0 In each of Problems 11 and 12, find the value of b for which the given equation is exact, and then solve it using that value of b. 11. (xy² + bx²y) + (x + y)x²y' = 0 12. (ye2xy + x) + bxe²xy y' = 0 13. Assume that equation (6) meets the requirements of Theorem 2.6.1 in a rectangle R and is therefore exact. Show that a possible function(x, y) is (x, y) = - fo where (xo, yo) is a point in R. M(s, yo) ds + Yo is also In eac exact factor 15. N(x, t) dt, 16. 17. only has In the 18 19 24 2
Transcribed Image Text:4. dy dx ax + by bx + cy - by bx - cy ax- +(6y²-x²+3) y' = 0 5. 6. (yexy cos(2x)-2exy sin(2x)+2x)+(xexy cos(2x) - 3) y' = 0 7. (y/x+6x) + (lnx-2) y' = 0, x > 0 X y dy (x² + y²) 3/2 dx = 0 + 8. (x² + y2)3/2 In each of Problems 9 and 10, solve the given initial value problem and determine at least approximately where the solution is valid. 9. (2x - y) + (2y - x) y' = 0, y(1) = 3 10. (9x² + y − 1) − (4y − x) y' = 0,_y(1) = 0 In each of Problems 11 and 12, find the value of b for which the given equation is exact, and then solve it using that value of b. 11. (xy² + bx²y) + (x + y)x²y' = 0 12. (ye2xy + x) + bxe²xy y' = 0 13. Assume that equation (6) meets the requirements of Theorem 2.6.1 in a rectangle R and is therefore exact. Show that a possible function(x, y) is (x, y) = - fo where (xo, yo) is a point in R. M(s, yo) ds + Yo is also In eac exact factor 15. N(x, t) dt, 16. 17. only has In the 18 19 24 2
In Example I it was relatively easy to see that the differential equation was exact and, in fact, easy to find its solution, at least implicitly, by recognizing the required function . For more complicated equations it may not be possible to do this so easily. How can we tell whether a given equation is exact, and if it is, how can we find the function (x, y)? The following theorem answers the first question, and its proof provides a way of answering the second. Theorem 2.6.1 Let the functions M, N, My, and Nx, where subscripts denote partial derivatives, be continuous in the rectangular¹9 region R: a < x < 3,7 < y < 6. Then equation (6) M(x, y) + N(x, y) y' = 0 is an exact differential equation in R if and only if My(x, y) = Nx(x, y) at each point of R. That is, there exists a function x(x, y) = M(x, y), if and only if M and N satisfy equation (10). satisfying equations (7), y(x, y) = N(x, y), (10) The proof of this theorem has two parts. First, we show that if there is a function such that equations (7) are true, then it follows that equation (10) is satisfied. Computing My and Nx from equations (7), we obtain yx(x, y). (11) yx are also continuous. This My(x, y) = xy(x, y), Nx(x, y) = Since My and Nx are continuous, it follows that xy and guarantees their equality, and equation (10) is valid. We now show that if M and N satisfy equation (10), then equation (6) is exact. The proof involves the construction of a function satisfying equations (7) x(x, y) = M(x, y), y(x, y) = N(x, y). While a complete discussion of when y(x, y) = c defines y = (x) implicitly as a differentiable function of x is eyond the scope and focus of this course, in general terms this condition is satisfied, locally, at points (x, y), where /dy(x, y) #0. More details can be found in most books on advanced calculus. It is not essential that the region be rectangular, only that it be simply connected. In two dimensions this means that e region has no holes in its interior. Thus, for example, rectangular or circular regions are simply connected, but an nular region is not. More details can be found in most books on advanced calculus.
Transcribed Image Text:In Example I it was relatively easy to see that the differential equation was exact and, in fact, easy to find its solution, at least implicitly, by recognizing the required function . For more complicated equations it may not be possible to do this so easily. How can we tell whether a given equation is exact, and if it is, how can we find the function (x, y)? The following theorem answers the first question, and its proof provides a way of answering the second. Theorem 2.6.1 Let the functions M, N, My, and Nx, where subscripts denote partial derivatives, be continuous in the rectangular¹9 region R: a < x < 3,7 < y < 6. Then equation (6) M(x, y) + N(x, y) y' = 0 is an exact differential equation in R if and only if My(x, y) = Nx(x, y) at each point of R. That is, there exists a function x(x, y) = M(x, y), if and only if M and N satisfy equation (10). satisfying equations (7), y(x, y) = N(x, y), (10) The proof of this theorem has two parts. First, we show that if there is a function such that equations (7) are true, then it follows that equation (10) is satisfied. Computing My and Nx from equations (7), we obtain yx(x, y). (11) yx are also continuous. This My(x, y) = xy(x, y), Nx(x, y) = Since My and Nx are continuous, it follows that xy and guarantees their equality, and equation (10) is valid. We now show that if M and N satisfy equation (10), then equation (6) is exact. The proof involves the construction of a function satisfying equations (7) x(x, y) = M(x, y), y(x, y) = N(x, y). While a complete discussion of when y(x, y) = c defines y = (x) implicitly as a differentiable function of x is eyond the scope and focus of this course, in general terms this condition is satisfied, locally, at points (x, y), where /dy(x, y) #0. More details can be found in most books on advanced calculus. It is not essential that the region be rectangular, only that it be simply connected. In two dimensions this means that e region has no holes in its interior. Thus, for example, rectangular or circular regions are simply connected, but an nular region is not. More details can be found in most books on advanced calculus.
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