This preview shows pages 1–4. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 ) 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 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 + c. Then, much as before, y + 2 = exp ( x + 1) 2 2 + c ! = e 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 ....
View Full
Document
 Spring '10
 ROBINSON
 Equations

Click to edit the document details