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Unformatted text preview: 1.6: Linear First Order Equations The Extended Chain Rule Recall that if y = H x , v H x LL , then dy dx = x dx dx + v dv dx = x + 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 4 Sqrt @ v D + 3 L , v D 1 8 J 3 + 4 v 3 Log B 3 + 4 v FN 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. We show the direction field along with the domain restriction. theDE = Sqrt @ 3 x + 4 y + 7 D ; ivp = 88 0, 1 < , 8 0, 1 < , 8 1.5, 1 < , 8 2, 2 < , 8 3.5, < , 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 4 x 7 4, 8 x, 4, 4 < , PlotStyle fi 8 Red, Thick <D D 4 2 2 4 3 2 1 1 2 3 x y Direction Field 2 Chapter1Section6.nb 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! 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 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 + 2 y @ 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 2 ProductLog 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 theDE = y H x + 2 y L ; ivp = 88 0, 1 2 < , 8 0, 1 < , 8 2, 2 < , 8 2, 2 < , 8 3, 1 << ; fig1 = 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 2 4 3 2 1 1...
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This note was uploaded on 01/13/2012 for the course MATH 306 taught by Professor Keithemmert during the Spring '11 term at Tarleton.
 Spring '11
 KeithEmmert
 Equations, Chain Rule

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