Chapter1Section6 - 1.6 Linear First Order Equations The...

This preview shows page 1 - 7 out of 20 pages.

1.6: Linear First Order Equations The Extended Chain Rule Recall that if y = Β H x , v H x LL , then dy dx = ¶Β x dx dx + ¶Β v dv dx = Β x + Β v dv dx . Form I: dy dx = F H a x + b y + c L . Substitution: v = a x + b y + c . Integrate @ 1 H 4Sqrt @ v D + 3 L , v D 1 8 J 3 + 4 v - 3Log B 3 + 4 v FN ª Example: Consider the differential equation dy dx = 3 x + 4 y + 7 . A. Solve the differential equation. B. List any restrictions for the solution to be meaningful.
Image of page 1

Subscribe to view the full document.

We show the direction field along with the domain restriction. theDE = Sqrt @ 3x + 4y + 7 D ; ivp = 88 0, 1 < , 8 0, - 1 < , 8 - 1.5, 1 < , 8 2, 2 < , 8 3.5, 0 < , 8 - 3, 2 << ; Show @ VectorPlot @8 1, theDE < , 8 x, - 4, 4 < , 8 y, - 3, 3 < , FrameLabel fi 8 x, y < , Axes fi True, VectorScale fi 8 Small, Tiny, None < , VectorStyle fi Gray, StreamScale fi Full, StreamStyle fi 8 Blue, Thick, "Line" < , StreamPoints fi 8 ivp < , PlotLabel fi "Direction Field", Epilog fi 8 PointSize @ Large D , Point @ ivp D< D , Plot @ - 3 4x - 7 4, 8 x, - 4, 4 < , PlotStyle fi 8 Red, Dashed, Thick <D D - 4 - 2 0 2 4 - 3 - 2 - 1 0 1 2 3 x y Direction Field 2 Chapter1Section6.nb
Image of page 2
Definition A h o m o g e n e o u s first order differential equation can be written in the form dy dx = F J y x N . Form II: Homogeneous DEs. Substitution: v = y x . Notice that when v = y x , we have y = xv, and hence dy dx = v + x dv dx since v is a function of x . Hence dy dx = F J y x N v + x dv dx = F H v L x dv dx = F H v L - v , Notice that the above is a separable equation! ª Example Consider the initial value problem H x + 2 y L y ' = y , y H 2 L = 1. A. Solve the initial value problem. B. List any restrictions. Chapter1Section6.nb 3
Image of page 3

Subscribe to view the full document.

Look! DSolve threatens to coughs up a hairball! It is warning us that inverse functions were used on something that may not be a function! sol1 = y @ x D . DSolve @8H x + 2y @ x DL y' @ x D y @ x D , y @ 2 D 1 < , y @ x D , x D InverseFunction::ifun: Inverse functions are being used. Values may be lost for multivalued inverses. Solve::ifun: Inverse functions are being used by Solve, so some solutions may not be found; use Reduce for complete solution information. Solve::ifun: Inverse functions are being used by Solve, so some solutions may not be found; use Reduce for complete solution information. : x 2ProductLog A ª x 2 E > Anyone heard of the ProductLog function? it is the principal solution for w in z = we w . If z ? - 1 e , then you have real solutions. 4 Chapter1Section6.nb
Image of page 4
theDE = y H x + 2y L ; ivp = 88 0, 1 2 < , 8 0, - 1 < , 8 2, 2 < , 8 - 2, 2 < , 8 3, - 1 << ; Show @ VectorPlot @8 1, theDE < , 8 x, - 4, 4 < , 8 y, - 3, 3 < , FrameLabel fi 8 x, y < , Axes fi True, VectorScale fi 8 Small, Tiny, None < , VectorStyle fi Gray, StreamScale fi Full, StreamStyle fi 8 Blue, Thick, "Line" < , StreamPoints fi 8 ivp < , PlotLabel fi "Direction Field", Epilog fi 8 PointSize @ Large D , Point @ ivp D< D , Plot @ sol1, 8 x, - 4, 4 < , PlotStyle fi 8 Red, Dashed, Thick <D D - 4 - 2 0 2 4 - 3 - 2 - 1 0 1 2 3 x y Direction Field Chapter1Section6.nb 5
Image of page 5

Subscribe to view the full document.

A safer way to plot the solution is to use ContourPlot after we solve it by hand. Notice that near the origin, the red dashed curve loops around on itself...this is NOT a function!
Image of page 6
Image of page 7

{[ snackBarMessage ]}

Get FREE access by uploading your study materials

Upload your study materials now and get free access to over 25 million documents.

Upload now for FREE access Or pay now for instant access
Christopher Reinemann
"Before using Course Hero my grade was at 78%. By the end of the semester my grade was at 90%. I could not have done it without all the class material I found."
— Christopher R., University of Rhode Island '15, Course Hero Intern

Ask a question for free

What students are saying

  • Left Quote Icon

    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.

    Student Picture

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

  • Left Quote Icon

    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.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    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.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern