m2k_dfq_sepfrd

# m2k_dfq_sepfrd - Separable First Order Dierential Equations...

This preview shows pages 1–4. Sign up to view the full content.

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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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 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 ) .

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern