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Unformatted text preview: 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 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: Consider dy dx + P H x L y = Q H x L , y H x L = y , where P H x L and Q H x L are both continuous on some interval I. Then the initial value problem has a unique solution given by y H x L = e P H x L dx B Q H x L e P H x L dx dx + C F H for an appropriate choice of C L which is defined on the entire interval I . Example Solve the differential equation y ' 2 xy = e x 2 . 2 Chapter1Section5.nb Of course, Mathematica can help with integrating factors... In[31]:= Integrate @ Exp @ 2 x D , x D Out[31]= 1 2  2 x Or just solve it... In[33]:= DSolve @ y' @ x D 2 x y @ x D Exp @ x^2 D , y @ x D , x D DSolve @8 y' @ x D 2 x y @ x D Exp @ x^2 D , y @ D 2 < , y @ x D , x D Out[33]= 99 y @ x D fi x 2 x + x 2 C @ 1 D== Out[34]= 99 y @ x D fi x 2 H 2 + x L== Chapter1Section5.nb 3 In[41]:= theDE = 2 x y + 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,...
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 Fall '11
 KeithEmmert
 Equations

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