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· y(x + y)dx + (x + 2y 1)ay (determination of integrating factor)

· y(x + y)dx + (x + 2y 1)ay (determination of integrating factor)

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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Can you solve this using the flow given for determination of integrating factor

Determination of Integrating Factor We have already mentioned in the previous lesson the aid of integrating factor in solving a first-order linear differential equation and that the same idea sometimes works for a nonexact differential equation M(x, y)dx + N(x, y)dy = 0 (1). We mean to say that it is sometimes possible to determine an integrating factor u(x, y) such that after multiplying it to a given nonexact equation (1), the left side of the resulting equation u(x, y) M(x, y)dx + u(x, y)N(x, y)dy = 0 (2) becomes an exact differential. To determine such integrating factor u(x, y), recall the criterion for exactness. Equation (2) is exact if and only if or U U Ə - (UM) = U ду Solving for u, u(u) = ef ƏM Əy ƏM ƏN dy дх + M ƏN əx du ƏN dy ?x du əx du U = a da (UN) = U + N = N If u satisfies (3), then u is an integrating factor for (1). Note that M, N, ƏN are dy əx known functions of r and y. To determine such function u, from (3), we have to solve for a partial differential equation, but were not yet ready to do that. Thus, we first make an assumption that u is a function of only one variable. Let us first assume that u depends on a alone. Then, = becomes Ju du ?х dx ƏM Ndu dr [ du = 12 (M u 1 (ƏM ƏN N dy Ər du əx - M ƏM ƏN Əy ?х ON N ду дх Solving for u, we have u(x) = Similarly, if u depends on y alone, then we will have, du ду It is still not easy to determine u even after taking the integral of both sides of (4) if its right side depends on both x and y. However, if after some algebraic simplications are made, 1 (OM ON the expression turns out to be dependent alone on the variable y, then (4) is now a first-order ordinary differential equation and we can finally solved for u in (4) by separation of variables. That is, ду Әх ƏM ƏN L (OM_W) dr dx du 1 /OM ƏN M ду əx dy du dy dx (3) dx =-M 1 ƏN -- (-2) dy U M Əy Əx ƏM , and Ju and = 0. Hence, (3) Əy
Transcribed Image Text:Determination of Integrating Factor We have already mentioned in the previous lesson the aid of integrating factor in solving a first-order linear differential equation and that the same idea sometimes works for a nonexact differential equation M(x, y)dx + N(x, y)dy = 0 (1). We mean to say that it is sometimes possible to determine an integrating factor u(x, y) such that after multiplying it to a given nonexact equation (1), the left side of the resulting equation u(x, y) M(x, y)dx + u(x, y)N(x, y)dy = 0 (2) becomes an exact differential. To determine such integrating factor u(x, y), recall the criterion for exactness. Equation (2) is exact if and only if or U U Ə - (UM) = U ду Solving for u, u(u) = ef ƏM Əy ƏM ƏN dy дх + M ƏN əx du ƏN dy ?x du əx du U = a da (UN) = U + N = N If u satisfies (3), then u is an integrating factor for (1). Note that M, N, ƏN are dy əx known functions of r and y. To determine such function u, from (3), we have to solve for a partial differential equation, but were not yet ready to do that. Thus, we first make an assumption that u is a function of only one variable. Let us first assume that u depends on a alone. Then, = becomes Ju du ?х dx ƏM Ndu dr [ du = 12 (M u 1 (ƏM ƏN N dy Ər du əx - M ƏM ƏN Əy ?х ON N ду дх Solving for u, we have u(x) = Similarly, if u depends on y alone, then we will have, du ду It is still not easy to determine u even after taking the integral of both sides of (4) if its right side depends on both x and y. However, if after some algebraic simplications are made, 1 (OM ON the expression turns out to be dependent alone on the variable y, then (4) is now a first-order ordinary differential equation and we can finally solved for u in (4) by separation of variables. That is, ду Әх ƏM ƏN L (OM_W) dr dx du 1 /OM ƏN M ду əx dy du dy dx (3) dx =-M 1 ƏN -- (-2) dy U M Əy Əx ƏM , and Ju and = 0. Hence, (3) Əy
5. y(x + y)dx + (x+2y-1)dy = 0 (determination of integrating factor)
Transcribed Image Text:5. y(x + y)dx + (x+2y-1)dy = 0 (determination of integrating factor)
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