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Introdu tion In this hapter, we will onsider ordinary dierential equations of rst order, that is, F(t;y;y0)=0: Unfortunately,onlyvery few spe ial types of rst order dierential equations admit solutions that an be expressed in terms of elementary fun tions. In fa t, one ould say that almost all rst order dierential equations an not be expressed. In the following, we will dis uss several spe ial lasses of rst order ODE that are solvable, illustrated by \textbook" examples, as well as some general existen e and uniqueness results. 1 Linear First Order ODE In this se tion we will fo us on linear rst order ODE of following type: dy dt +p(t)y = g(t); sometime y_ +p(t)y = g(t): This lass of dierential equations are easy to solve. Indeed, here we have our rst existen e and uniqueness result: Theorem: Provided p(t) and g(t) are ontinuous fun tions on an open interval I =(;), there exists a unique solution to the initial value problem dy +p(t)y = g(t); y(t )=y dt 0 0 on whole interval I,forevery t 2 I and y 2 IR. A tually, the solution is given by 0 0 Rt (s)g(s)ds+y0 4 Rt p(s)ds t0 t y(t)= ; where (t) = e 0 ; 8 t 2 I: (t) Idea of the proof: The tri k of solving this dierential equation is the following: Suppose we an nd a fun tion (t) su h that _ (t)=(t)p(t): It follows that, bymultiplying (t) on both sides of the equation, (t)_y + (t)p(t)y = (t)g(t) ) (t)_y +_(t)y = d [(t)yâ„„=(t)g(t) dt 1 whi h yields (t)y = Z (t)g(t)dt + ) y = R (t)g(t)dt + : (t) However, su h (t) alwasy exists as _ (t) d R p(t)dt+ (t) = dt ln(t)=p(t) ) (t)=e : All su h fun tions diers up to a multipli ative fa tor. A possible hoi e is Rt p(s)ds Rt (s)g(s)ds +y0 t t0 (t)=e 0 ) y= ; (t) whi h is a solution to the dierential equation. Proof: Existen e: It is straight-forward to verify that fun tion y dened above is a solution to the dierential equation. Uniqueness: Suppose both y (t) and y (t) are solutions to the dierential equations. That is 1 2 y0 +p(t)y = g(t); y (t )=y ; 8 i =1;2 i i i 0 0 4 Dene z(t) = y (t) y (t). It follows that 1 2 z0 + p(t)z =0; z(t0)=0; Multiplying both side by (t)we obtain 0 d (t)z +(t)p(t)z = dt[(t)zâ„„=0; whi h implies that (t)z(t) onst. In parti ular, (t)z(t) (t0)z(t0)=0.It follows that z(t) 0 sin e (t) > 0. Therefore y (t) y (t). 2. 1 2 Corollary: Suppose p(t) r 6=0,g(t) b are both onstant. The dierential equation y0 = ry+b; y(t0)=y0; has a unique solution b r(t t0) b y(t)= y e + : 0 r r Remark: Here the existen e and unique result is a global result, sin e the solution we nd is dened on the whole interval. Later we will see that a lo al existen e and uniqueness result for non-linear rst order dierential equation. Remark: The multipli ator (t) is alled integrating fa tor. 2 Example (Comparison Prin iple): Suppose x is a solution to the initial value problem dx +p(t)x = g (t); x(t )=x dt 1 0 0 while y is a solution to the initial value problem dy +p(t)y = g (t); y(t )=y : dt 2 0 0 If g (t) g (t); 8t t and x y , then x(t) y(t) for all t t . 1 2 0 0 0 0 Proof: Without loss of generality, assume t =0.Dene z(t)=x(t) y(t). It is easy to see 0 that z satises dierential equation z0 +p(t)z = g1(t) g2(t): R p(t)dt Multiplying both sides byintegrating fa tor = e ,wehave d (t)z +p(t)(t)z = dt (t)z = (t)(g1 g2) 0 This implies that (t)z(t) is a non-de reasing fun tion. In parti ular, (t)z(t) (0)z(0) 0 ) z(t)0 for all t. This ompletes the proof. 2 Exer ise: (Comparsion Prin iple) Following the pre eding example, but assume x(t) is instead a solution to a general the initial value problem dx dt = f(t;x); x(t0)=x0 where f(t;x) is a ontinuous fun tion su h that f(t;x) p(t)x + g2(t) for all x and all t t0. Show the omparison inequality still holds, i.e. x(t) y(t) for all t t0. Example: 1. Solve the following dierential equation y0 + 1y =1; 8 x>0: x 2. If wehave initial ondition y(1) = 1, determine y. Solution: 1. The integrating fa tor (x) is determined by 0 (x) d 1 (x)= x ) dxln= x ) ln(x)=lnx+ ) (x)= x: Any hoi e of will yield a integrating fa tor, and any one will do. We just pi k =1 and wehave 0 0 1 2 x xy +y =x ) (xy) =x ) xy= 2x + ) y= 2 + x: 3 1 1 1 2. y(1) = 1 yields = 2. Hen e y(x)=2 x+ x . 2 Exer ise: Solve the following dierential equation xy0 +2y = x4 +1; x>0 Solution: Multiply both sides by (x)=x,we obtain that 6 2 2 0 5 2 0 5 2 x x x y +2xy =x +x ) (x y) =x +x ) x y = 6 + 2 + Therefore the general solution of the equation is 4 y = x + 1 + : 6 2 x2 Example (A geometri problem) Find the family of urves with the property that the segmentof a tangent line drawn between a point of tangen y and Y-axis is bise ted bytheX-axis. Solution: Suppose P =(x;y)isapoint on the urve, and the tangent line drawn from point P will interse t Y -axis at point Q whose oordinates are, say (0;z). Sin e the mid-pointof P and Q has oordinates x+0 y+z ; 2 2 wemust have z = y,whi h implies that dy =slope of tangentline= y y = 2y: dx 0 x x Multiplying integrating fa tor (x)= 1 ,wehave x2 d y =0 ) y= x2 for some onstant : 2 dx x The family of urves we are looking for is y = x2. 2 Exer ise: Find the family of urves su h that the tangent line drawn between X-axis and Y-axis is bise ted by the point of onta t. That is, point P is the mid-pointofQ and R. 4
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