Chapter1Section6

# Chapter1Section6 - 1.6 Linear First Order Equations The...

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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.

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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
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

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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
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

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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!

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