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Unformatted text preview: Separable First Order Differential Equations Form of Separable Equations These are differential equations which take the form dy dx = g ( x ) h ( y ) or dy dx = g ( x ) ˆ h ( y ) , where g ( x ) is a continuous function of x and h ( y ) is a continuously differentiable function of y (to guarantee unique solutions). The two forms agree if ˆ h ( y ) = 1 /h ( y ). In differential form we can rewrite such an equation as h ( y ) dy = g ( x ) dx thus separating the y dependence from the x dependence. We have already encountered the simplest example, i.e., the homogeneous first order linear equation dy dx + p ( x ) y = 0 . We can rewrite this in the form 1 y dy = p ( x ) dx. Method of Solution Integrating both sides of h ( y ) dy = g ( x ) dx we have Z y h ( r ) dr = Z x g ( s ) ds + c, where c is an arbitrary constant. Then, letting H ( y ) = R y h ( r ) dr and G ( x ) = R x g ( s ) ds , we have H ( y ) = G ( x ) + c or H ( y ) G ( x ) c = 0 . This is a parametric equation for y in terms of x ; if it can be solved for y to give, explicitly, y = y ( x, c ) = H 1 ( G ( x ) + c ) , 1 then we have an explicit formula for the general solution. Example 1 Consider the differential equation dy dx = xy + 2 x + y + 2 . Here we can factor the right hand side: xy + 2 x + y + 2 = ( x + 1)( y + 2). So we have 1 y + 2 dy = ( x + 1) dx and, on integration, we have log  y + 2  = ( x + 1) 2 2 + ˆ c. Then, much as before, y + 2 = ± exp ( x + 1) 2 2 + ˆ c ! = ± e ˆ c exp ( x + 1) 2 2 ! = c exp ( x + 1) 2 2 ! so that the general solution becomes y ( x, c ) = c exp( ( x + 1) 2 2 ) 2 . Example 2 Consider the differential equation dy dx = y 2 x 2 + 3 x + 2 . This becomes 1 y 2 dy = 1 x 2 + 3 x + 2 dx. At this point we need to pause and provide a 2 Reminder: Partial Fractions Decomposition When we need to integrate a function which is a quotient of two polynomials, i.e., f ( x ) = b 1 x n 1 + b 2 x n 2 + ··· + b n a x n + a 1 x n 1 + ··· a n 1 x + a n , a 6 = 0 , we first need to transform f ( x ) into another form. This is done by the procedure of partial fractions decomposition . The first step is to factor the denominator. Assuming it has distinct roots r 1 , r 2 , ··· r n the denominator takes the form a ( x r 1 )( x r 2 ) ··· ( x r n ). We then try to write f ( x ) = c 1 x r 1 + c 2 x r 2 + ··· + c n x r n ....
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This note was uploaded on 01/23/2012 for the course MATH 4254 taught by Professor Robinson during the Spring '10 term at Virginia Tech.
 Spring '10
 ROBINSON
 Equations

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