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Unformatted text preview: Homework 1  Due on January 22 (Thursday)  Solution 1 1. Find the general solutions of the following 1st order ODE: (a) xy + 2 y = sin x First, find the integrating factor μ ( x ): xμ ( x ) y + 2 μ ( x ) y = μ ( x )sin x . [ xμ ( x )] = 2 μ ( x ) ⇒ μ ( x ) + xμ ( x ) = 2 μ ( x ) ⇒ xμ ( x ) = μ ( x ) ⇒ μ ( x ) /μ ( x ) = 1 /x ⇒ [ln  μ ( x )  ] = 1 /x ⇒ ln  μ ( x )  = ln  x  ⇒ μ ( x ) = x Thus we have x 2 y + 2 xy = [ x 2 y ( x )] = x sin x . Hence, integrating by parts, we get x 2 y ( x ) = Z x sin xdx = x ( cos x ) Z ( cos x ) dx = x cos x +sin x + c. So the general solution is y ( x ) = cos x x + sin x x 2 + c x 2 . (b) (1 + x 2 ) y + 4 xy = 1 1 + x 2 First, find the integrating factor μ ( x ): (1 + x 2 ) μ ( x ) y + 4 xμ ( x ) y = μ ( x ) 1 + x 2 . £ (1 + x 2 ) μ ( x ) / = 4 xμ ( x ) ⇒ 2 xμ ( x ) + (1 + x 2 ) μ ( x ) = 4 xμ ( x ) ⇒ (1 + x 2 ) μ ( x ) = 2 xμ ( x ) ⇒ μ ( x ) /μ ( x ) = 2 x/ (1 + x 2 ) ⇒ [ln  μ ( x )  ] = [ln  1 + x 2  ] ⇒ ln  μ ( x )  = ln  1 + x 2  ⇒ μ ( x ) = 1 + x 2 Thus we have (1 + x 2 ) 2 y + 4 x (1 + x 2 ) y = [(1 + x 2 ) 2 y ] = 1. Hence, by integrating, we get (1 + x 2 ) 2 y ( x ) = x + c . So the general solution is y ( x ) = x + c (1 + x 2 ) 2 . MATH2400  Introduction to Differential Equations  Spring 2004  JeongRock Yoon Homework 1  Due on January 22 (Thursday)  Solution 2 2. Solve the following initial value problems:2....
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This homework help was uploaded on 02/24/2008 for the course MATH 2400 taught by Professor Yoon during the Spring '04 term at Rensselaer Polytechnic Institute.
 Spring '04
 Yoon
 Differential Equations, Equations

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