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Solve the equation 3uy + xy = 0. (Hint: Let v = Uy.)

Solve the equation 3uy + xy = 0. (Hint: Let v = Uy.)

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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[Partial Differential Equations] How do you solve number 2?

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2. Solve the equation 3uy + xy = = 0. (Hint: Let v = =Uy.)
Transcribed Image Text:2. Solve the equation 3uy + xy = = 0. (Hint: Let v = =Uy.)
Example 3. Solve the PDE ux + 2xy²uy = 0. (8) The characteristic curves satisfy the ODE dy/dx = = 2xy²/1 = 2xy². To solve the ODE, we separate variables: dy/y² = 2x dx; hence -1/y = x² - C, so that y = (C - x²) ¹ (9) These curves are the characteristics. Again, u(x, y) is a constant on each such curve. (Check it by writing it out.) So u(x, y) = f(C), where f is an arbitrary function. Therefore, the general solution of (8) is obtained by solving (9) for C. That is, u(x, y) ;, y) = f ( x ² + 1). (10) Again this is easily checked by differentiation, using the chain rule:ux = 2x. f'(x² + 1/y) and uy = −(1/y²) · f'(x² + 1/y), whence ux + 2xy²uy = 0. In summary, the geometric method works nicely for any PDE of the form a(x, y)ux + b(x, y)uy = 0. It reduces the solution of the PDE to the solution of the ODE dy/dx = b(x, y)/a(x, y). If the ODE can be solved, so can the PDE. Every solution of the PDE is constant on the solution curves of the ODE. Moral Solutions of PDEs generally depend on arbitrary functions (instead of arbitrary constants). You need an auxiliary condition if you want to deter- mine a unique solution. Such conditions are usually called initial or boundary conditions. We shall encounter these conditions throughout the book.
Transcribed Image Text:Example 3. Solve the PDE ux + 2xy²uy = 0. (8) The characteristic curves satisfy the ODE dy/dx = = 2xy²/1 = 2xy². To solve the ODE, we separate variables: dy/y² = 2x dx; hence -1/y = x² - C, so that y = (C - x²) ¹ (9) These curves are the characteristics. Again, u(x, y) is a constant on each such curve. (Check it by writing it out.) So u(x, y) = f(C), where f is an arbitrary function. Therefore, the general solution of (8) is obtained by solving (9) for C. That is, u(x, y) ;, y) = f ( x ² + 1). (10) Again this is easily checked by differentiation, using the chain rule:ux = 2x. f'(x² + 1/y) and uy = −(1/y²) · f'(x² + 1/y), whence ux + 2xy²uy = 0. In summary, the geometric method works nicely for any PDE of the form a(x, y)ux + b(x, y)uy = 0. It reduces the solution of the PDE to the solution of the ODE dy/dx = b(x, y)/a(x, y). If the ODE can be solved, so can the PDE. Every solution of the PDE is constant on the solution curves of the ODE. Moral Solutions of PDEs generally depend on arbitrary functions (instead of arbitrary constants). You need an auxiliary condition if you want to deter- mine a unique solution. Such conditions are usually called initial or boundary conditions. We shall encounter these conditions throughout the book.
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