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Unformatted text preview: AM 3815  Solutions to Assignment 1 1. Exercise 2, page 9, Strauss Following the hint, we substitute v = u y into the equation 3 u y + u xy = 0 to obtain 3 v + v x = 0 v x = 3 v . It follows that v = f ( y ) e 3 x = u y , and antidifferentiating this equation we find u = Z f ( y ) dy + g ( x ) e 3 x , and so the solution is u ( x , y ) = F ( y )e 3 x + G ( x ). 2. Exercise 5, page 10, Strauss We are asked to solve xu x + yu y = 0. To apply the method of characteristics we first solve the initial value problem, dx dt = x with x (0) = x dy dt = y with y (0) = y to obtain x ( t ) = x e t , y ( t ) = y e t . Now we define U ( t ) = u ( x ( t ), y ( t )). Differentiating with respect to t we find U ( t ) = x u x + y u y = 0 and so U ( t ) = something that is constant with respect to t . When we translate back to x , ycoordinates, then, we have u ( x , y ) = f , but f cannot depend on x and y in a way that would dependent on t . From above we see that y ( t ) / x ( t ) does not depend on t and so we can assert u ( x , y ) = f ( y / x ). 3. Exercise 6, page 10, Strauss 1 We can again apply the Method of Characteristics. We first solve the initial value problem dx dt = 1 x 2 with x (0) = x dy dt = 1 with y (0) = y to get x ( t ) = sin( t + x ), y ( t ) = t + y and so if we combine x and y as ( y y ) (arcsin x x ) = ( y arcsin x ) + ( x y ) we get a composite variable, namely ( y arcsin x ), that is not dependent on t ....
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