m2k_dfq_sepfrd - Separable First Order Dierential Equations...

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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 re-write 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 re-write 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
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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
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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 0 x n + a 1 x n - 1 + · · · a n - 1 x + a n , a 0 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 0 ( 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 . Recombining the right hand side into a single fraction with a common denominator we have f ( x ) = c 1 ( x - r 2 ) · · · ( x - r n ) + · · · + c n ( x - r 1 ) · · · ( x - r n - 1 ( x - r 1 )( x - r 2 ) · · · ( x - r n ) .
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