smt1 - polynomial 2-2 + 1 = 0 of multiplicity 2. Therefore,...

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Math 5525. February 24, 2010. Midterm Exam 1. Problems and Solutions. Problem 1. Find the general solution of the equation dy dx - xy 2 = xy. Solution. We have dy dx = x ( y 2 + y ) = dy y ( y + 1) = xdx = ± 1 y - 1 y + 1 ² dy = xdx = ln | y | - ln | y + 1 | = 1 2 x 2 + C 0 = 1 + 1 y = y + 1 y = Ce - 1 2 x 2 , and y 0 . Problem 2. Find the solution of the equation x dy dx = 2 y + 3 x satisfying the initial condition y (1) = 0. Solution. This is a linear equation. The corresponding linear homogeneous equation xy 0 = 2 y has a solution y 1 = x 2 . One can find solutions of the given equation in the form y = C ( x ) y 1 = C ( x ) · x 2 . Then x ( C 0 x 2 + C · 2 x ) = 2 Cx 2 + 3 x - 1 , C 0 = 3 x - 4 , C ( x ) = C 1 - x - 3 , and y = C ( x ) · x 2 = C 1 x 2 - x - 1 . Since y (1) = 0, we have C 1 = 1, and y = x 2 - x - 1 . Problem 3. Find the general solution of the equation x dy dx = y (ln y - ln x ) . Solution. This is a homogeneous equation. dy dx = y x ln y x , u = y x = u + x du dx = u ln u, du u (ln u - 1) = dx x , ln | ln u - 1 | = ln | x | + C 0 , ln u - 1 = Cx, y = xu = xe Cx +1 . Problem 4. Find the solution of the equation y 00 - 2 y 0 + y = xe x satisfying the initial conditions y (0) = y 0 (0) = 0. Solution. Note that the exponent in the right side has coefficient 1, which is the root of the characteristic
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Unformatted text preview: polynomial 2-2 + 1 = 0 of multiplicity 2. Therefore, we can nd a particular solution y in the form y = x 2 ( Ax + B ) e x . Then Ly = ( D-1) 2 e x ( Ax 3 + Bx 2 ) = e x D 2 ( Ax 3 + Bx 2 ) = e x (6 Ax + 2 B ) = e x x, which implies A = 1 / 6 , B = 0, and y = x 3 6 e x . This solution satises the conditions y (0) = y (0) = 0. Problem 5. Find the general solution of the equation ( y ) 2-2 yy 00 = 0 . Solution. Substituting y ( x ) = z ( y ), we get y 00 ( x ) = zz , z 2 = 2 yzz , 2 dz z = dy y , 2 ln | z | = ln | y | = A 1 , z 2 = A 2 y, z = dy dx = A 3 y 1 / 2 , y-1 / 2 dy = A 3 dx, y 1 / 2 = C 1 x + C 2 , y = ( C 1 x + C 2 ) 2 ....
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