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Unformatted text preview: Math 2a Practical, Fall 2008: Solutions to Homework 3 You are bound by the honor code to not look at this file if you did not take Math 2a in the Fall of 2008 1. (Bonus questions for practice) Find all solutions of the following differential equations. Note that it may help to make good substitutions! (20pts) (a) y = xy +cos( x ) xy x 2 Solution. Let’s observe first that any solution of this differential equation must be defined on an interval which does not contain 0. We can rewrite the given differential equation in the form ( xy x 2 ) y = xy + cos x or equivalently xy cos x + ( xy x 2 ) y = 0 ( * ) or M ( x,y ) + N ( x,y ) y = 0 for M ( x,y ) = xy cos x and N ( x,y ) = ( xy x 2 ) . Let’s observe that M y N x N = x y xy x 2 = 1 x so using the same method as in example 4 from page 99 in the textbook (eighth edition) we get that an integrating factor μ of the given differential equation is a solution of the differential equation dμ dx = M y N x N μ = μ x A solution of this differential equation (with separable variables) is μ ( x ) = 1 x Multiplying the equation (*) by this integrating factor we obtain the equation ( y + cos x x ) + ( x y ) y = 0. Thus there is a ψ ( x,y ) such that ψ x ( x,y ) = y + cos x x , ψ y ( x,y ) = ( x y ) Integrating the first of these equations we get ψ ( x,y ) = xy + R cos x x + h ( y ) and from the second equation we get x + h ( y ) = x y so h ( y ) = y so h ( y ) = y 2 2 (we can omit the constant). Therefore we get ψ ( x,y ) = xy + R cos x x y 2 2 and therefore the solutions of the given differential equation are given implicitly by xy + Z cos x x y 2 2 = 0 Solving the quadratic equation we get y ( x ) = x ± s x 2 + 2 Z cos x x . To make the arbitrary constant explicit, we can write y ( x ) = x ± s x 2 + 2 Z x 1 cos x x dx + c, 1 if x > 0 or y ( x ) = x ± s x 2 + 2 Z x 1 cos x x dx + c, if x < 0. The existence interval of x , not easily computed, might be smaller. (b) y = 1 + y x + ( y x ) 2 Solution. This is a homogeneous equation as introduced in page 49. (This homogeneity of firstorder differential equations is different from the homogeneity of second or higherorder linear differential equations.) To solve a homogeneous equation we introduce a new dependent variable v so that v = y/x . Then y = dy/dx = d ( vx ) /dx = v + x dv/dx . Then the original equation is v + x dv/dx = 1 + v + v 2 , or x dv/dx = 1 + v 2 . This is changed to the separable equation: dv 1 + v 2 = dx x . Integrating, we obtain arctan v = ln  x  + c for some constant c . Replacing v by y/x , we find arctan( y/x ) = ln  x  + c , or y = x (tan(ln  x  + c )) . This solution is welldefined as long as x 6 = 0 and ln  x  + c 6 = kπ + π/ 2, i.e., x 6 = ± e kπ + π/ 2 c for any integer k ....
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This note was uploaded on 01/08/2011 for the course MA 2a taught by Professor Makarov,n during the Fall '08 term at Caltech.
 Fall '08
 Makarov,N
 Differential Equations, Statistics, Equations, Probability

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