# midsol1 - MAP 2302 Section 0100 Fall-2011 Midterm 1 1 Solve...

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MAP 2302 Section 0100 Fall-2011 Midterm 1 1. Solve the equation (1 - x 2 ) dy dx = 2 xy + 3 . 1st SOLUTION: This equation is linear. It has the following standard form y 0 - 2 x 1 - x 2 y = 3 1 - x 2 . Then μ = e R P = e - R 2 x 1 - x 2 dx = e ln 1 - x 2 = 1 - x 2 . Then ( y (1 - x 2 )) 0 = 3 and y (1 - x 2 ) = 3 x + C . It can be solved for y : y = 3 x + C 1 - x 2 . 2nd SOLUTION: This equation is exact: (2 xy + 3) dx + ( x 2 - 1) dy = 0 with ∂M ∂y = 2 x = ∂N ∂x . Then R Mdx = x 2 y + 3 x + g ( y ) with g 0 ( y ) = N - ( x 2 y + 3 x ) ∂y = x 2 - 1 - x 2 = - 1 . Thus, g ( y ) = - y and x 2 y + 3 x - y = C is the solution. Clearly that it can be written as y = 3 x + C 1 - x 2 . 2. Is the equation ( x - y ) dx + xdy = 0 exact? Justify your answer. Solve it? SOLUTION: No, since ( x - y ) ∂y = - 1 6 = 1 = ∂x ∂x . 1

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It is a homogeneous equation dy dx = y x - 1. With the substitution v = y x we obtain x dv dx + v = v - 1. Thus, dv dx = - 1 x . Therefore, v = - ln | x | + C and finally, y = Cx - x ln | x | . It is also linear and it has a special integrating factor. 3. Solve the initial value problem y 00 - 4 y 0 + 4 y = 0; y (1) = 0 , y 0 (1) = 1 SOLUTION. The characteristic equation r 2 - 4 r + 4 = 0 has a double root r = 2. Thus, y = C 1 e
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