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Special Second Order Equations (Sect. 2.2). Special Second ...

Special Second Order Equations ( sect . ).ISpecial Second Order nonlinear (Simpler)IVariabletmissing. (Harder)IReduction Order Second Order nonlinear equationsDefinitionGiven a functionsf:R3 R,a Second orderdifferential equationin the unknown functiony:R Ris given byy =f(t,y,y ).The equation islinearifffis linear in the argumentsyandy .Remarks:INonlinear Second Order differential equation are usuallydifficult to , there are two particular cases wheresecond orderequations can be transformed intofirst orderequations.(a)y =f(t,y ).The functionyis missing.(b)y =f(y,y ).The independent variabletis Second Order Equations ( sect . ).ISpecial Second Order nonlinear (Simpler)IVariabletmissing. (Harder)IReduction Order Second Second Order differential equation has the formy =f(t,y ),then the equation for v=y is the first Order equationv =f(t,v).

Special Second Order Equations (Sect. 2.2). I Special Second order nonlinear equations. I Function y missing. (Simpler) I Variable t missing. (Harder) I Reduction order method. Special Second order nonlinear equations Definition Given a functions f : R3 → R, a second order differential equation in the unknown function y : R → R is given by

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Transcription of Special Second Order Equations (Sect. 2.2). Special Second ...

1 Special Second Order Equations ( sect . ).ISpecial Second Order nonlinear (Simpler)IVariabletmissing. (Harder)IReduction Order Second Order nonlinear equationsDefinitionGiven a functionsf:R3 R,a Second orderdifferential equationin the unknown functiony:R Ris given byy =f(t,y,y ).The equation islinearifffis linear in the argumentsyandy .Remarks:INonlinear Second Order differential equation are usuallydifficult to , there are two particular cases wheresecond orderequations can be transformed intofirst orderequations.(a)y =f(t,y ).The functionyis missing.(b)y =f(y,y ).The independent variabletis Second Order Equations ( sect . ).ISpecial Second Order nonlinear (Simpler)IVariabletmissing. (Harder)IReduction Order Second Second Order differential equation has the formy =f(t,y ),then the equation for v=y is the first Order equationv =f(t,v).

2 ExampleFindysolution of the Second Order nonlinear equationy = 2t(y )2with initial conditionsy(0) = 2,y (0) = :Introducev=y .Thenv =y ,andv = 2t v2 v v2= 2t 1v= t2+ ,1y =t2 c, that is,y =1t2 initial condition implies 1 =y (0)= 1c c= 1 y =1t2 Second theysolution of the Second Order nonlinear equationy = 2t(y )2with initial conditionsy(0) = 2,y (0) = :Then,y= dtt2 1+ Fractions!1t2 1=1(t 1)(t+ 1)=a(t 1)+b(t+ 1).Hence, 1 =a(t+ 1) +b(t 1).Evaluating att= 1 andt= 1we geta=12,b= 1=12[1(t 1) 1(t+ 1)].y=12(ln|t 1| ln|t+ 1|)+ =y(0)=12(0 0) + concludey=12(ln|t 1| ln|t+ 1|)+ Second Order Equations ( sect . ).ISpecial Second Order nonlinear (Simpler)IVariabletmissing. (Harder)IReduction Order Second a Second Order differential equationy =f(y,y ), andintroduce the functionv(t) =y (t). If the function y is invertible,then the new functionw(y) =v(t(y))satisfies the first orderdifferential equationdwdy=1wf(y,w(y)).

3 Proof:Denote w=dwdyandv = thatv (t) =f(y,v(t)).By chain rule w=dwdy y=dvdt t(y)dtdy t(y)=v y t(y)=v v t(y)=f(y,v)v t(y).Therefore, w=f(y,w) Second a solutionyto the Second Order equationy = 2y y .Solution:The variabletdoes not appear in the ,v(t) =y (t).The equation isv (t) = 2y(t)v(t).Now introducew(y) =v(t(y)).Then w=dwdy=(dvdtdtdy) t(y)=v y t(y)=v v t(y).Using the differential equation, w=2yvv t(y) w= 2y v(y) =y2+ (t) =w(y(t)),we getv(t) =y2(t) + Second a solutionyto the Second Order equationy = 2y y .Solution:Recall:v(t) =y2(t) + is a separable equation,y (t)y2(t) +c= we only need to find a solution of the equation, and theintegral depends on whetherc>0,c= 0,c<0, we choose (forno Special reason) only one case,c= 1. dy1 +y2= dt+c0 arctan(y) =t+c0y(t) = tan(t+c0).

4 Again, for no reason, we choosec0= 0,and we conclude that onepossible solution to our problem isy(t) = tan(t).CSpecial Second Order Equations ( sect . ).ISpecial Second Order nonlinear (Simpler)IVariabletmissing. (Harder)IReduction Order of the Order methodRemark:Knowing one solution to a differential equation is enoughto find a Second solution not proportional to the first a non-zero function y1is solution toy +p(t)y +q(t)y= 0.(1)where p, q are given functions, then a Second solution to this sameequation is given byy2(t) =y1(t) e P(t)y21(t)dt,(2)where P(t) = p(t)dt. Furthermore, y1and y2are fundamentalsolutions to Eq. (1).Reduction of the Order methodExampleFind a fundamental set of solutions tot2y + 2ty 2y= 0,knowing thaty1(t) =tis a :Expressy2(t) =v(t)y1(t).The equation forvcomesfromt2y 2+ 2ty 2 2y2= need to computey2=v t,y 2=t v +v,y 2=t v + 2v.

5 So, the equation forvis given byt2(t v + 2v )+ 2t(t v +v) 2t v= 0t3v + (2t2+ 2t2)v + (2t 2t)v= 0t3v + (4t2)v = 0 v +4tv = of the Order methodExampleFind a fundamental set of solutions tot2y + 2ty 2y= 0,knowing thaty1(t) =tis a :Recall:v +4tv = is a first Order equation forw=v ,given byw +4tw= 0,sow w= 4t ln(w) = 4 ln(t) +c0 w(t) =c1t 4,c1 obtainv,that is,v=c2t 3+c3, withc2,c3 thaty2=t vwe then conclude thaty2=c2t 2+ 1 andc3= 0 we obtain the fundamental solutionsy1(t) =tandy2(t) = of the Order methodProof of the Theorem:The choice ofy2=vy1impliesy 2=v y1+v y 1,y 2=v y1+ 2v y 1+v y information introduced into the differential equation says that(v y1+ 2v y 1+v y 1) +p(v y1+v y 1) +qv y1= 0y1v + (2y 1+p y1)v + (y 1+p y 1+q y1)v= functiony1is solution ofy 1+p y 1+q y1= , the equation forvis given byy1v + (2y 1+p y1)v = of the Order methodRecall:y1v + (2y 1+p y1)v = is a first Order eq.

6 Forw(t) =v (t).That is,w +(2y 1y1+p)w= is the origin of hte name:Reduction of Order factor: =y21eP,withP = (y21ePw) = 0 w=w0e P/y21choosew0= =e P/y21,hencev(t) = e Py21dt y2(t) =y1(t) e P(t)y21(t)dtReduction of the Order methodProof:Recally1v + (2y 1+p y1)v = now need to showthaty1andy2=vy1are linearly y1vy1y 1(v y1+vy 1) =y1(v y1+vy 1) vy1y obtainWy1y2=v we havew=v ,v =w=e P/y21 y21v =e PRecall thatPis a primitive ofp, that is,P (t) =p(t),then weobtainWy1y2=e P,which is conclude thaty1andy2=vy1are linearlyindependent.


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