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Chapter1Section5

# Chapter1Section5 - 1.5 Linear First Order Equations Basic...

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1.5: Linear First Order Equations Basic Definitions A linear first order equation has the form dy dx + P H x L y = Q H x L , where P H x L and Q H x L are both continuous on some interval I . An integrating factor is Μ H x L = e P H x L dx . The Theory Notice that d dx A y H x L e P H x L dx E = y ' H x L e P H x L dx + y H x L P H x L e P H x L dx = e P H x L dx Q H x L . So, we can solve liner first order equations by multiply- ing by an integrating factor, e P H x L dx . Hence, we have y H x L = e - P H x L dx A Q H x L e P H x L dx dx + C E . Theorem: Considerdy dx + P H x L y = Q H x L , y H x 0 L = y 0 , whereP H x L andQ H x L arebothcontinuousonsomeintervalI. Thentheinitialvalueproblemhasauniquesolutiongivenby y H x L = e - P H x L dx A Q H x L e P H x L dx dx + C E H foranappropriatechoiceof C L whichisdefinedontheentireinterval I .

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Example Solve the differential equation y ' - 2 xy = e x 2 . 2 Chapter1Section5.nb
Of course, Mathematica can help with integrating factors... Exp @ Integrate @ - 2x, x DD ª - x 2 Or just solve it... DSolve @ y' @ x D - 2xy @ x D Exp @ x^2 D , y @ x D , x D DSolve @8 y' @ x D - 2xy @ x D Exp @ x^2 D , y @ 0 D - 2 < , y @ x D , x D 99 y @ x D fi ª x 2 x + ª x 2 C @ 1 D== 99 y @ x D fi ª x 2 H - 2 + x L== Chapter1Section5.nb 3

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theDE = 2xy + Exp @ x^2 D ; ivp = 88 0, 0 < , 8 0, 1 < , 8 0, - 1 < , 8 2, 1 < , 8 1, - 2 < , 8 - 2, 1 < , 8 - 1, 2 << ; VectorPlot @8 1, theDE < , 8 x, - 2.1, 2.1 < , 8 y, - 4, 4 < , 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 - 2 - 1 0 1 2 - 4 - 2 0 2 4 x y Direction Field 4 Chapter1Section5.nb
Example

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Chapter1Section5 - 1.5 Linear First Order Equations Basic...

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