FIRST-ORDER ORDINARY DIFFERENTIAL EQUATIONS

1 FIRST-ORDER ORDINARY DIFFERENTIAL EQUATIONSG(x, y, y ) = 0 in normal form:y =F(x, y) in DIFFERENTIAL form:M(x, y)dx+N(x, y)dy= 0 Last time we discussed first-orderlinearODE:y +q(x)y=h(x).We next consider FIRST-ORDER ODEs No general method of solution for 1st- order ODEs beyond linear case;rather, a variety of techniques that work on a case-by-case :i) Bring equation to separated-variables form, that is,y = (x)/ (y);then equation can be covered by this includey = (ax+by);y = (y/x).ii) Reduce to linear equation by transformation of of this include Bernoulli s ) Bring equation to exact- DIFFERENTIAL form, that isM(x, y)dx+N(x, y)dy= 0such thatM= / x,N= / solution determined from (x, y) =const. Useful reference for the ODE part of this course(worked problems and examples)Schaum s Outline SeriesDifferential EquationsR. Bronson and G. CostaMcGraw-Hill (Third Edition, 2006) Chapters 1 to 7: FIRST-ORDER order nonlinear equationsAlthough no general method for solution is available, there are several cases of physically relevant nonlinear EQUATIONS which can be solved analytically :Separable equationsd ( )d ( )y f xx g y=( ) ( )g y dy f x dx=!

2 !Solution :2dedxyyx=Ex 12e dxdyxy=!" "1excy!= + (e )xyc!=+orAlmost separable equationsd( )dyf ax byx=+z ax by= +ddddyzxxa b= +Change variables :1d( ( ))xzabf z=.+!d( )dzabf zx= +!2d( 4 )dyx yx=!+4z y x=!2dd44ddzyzxx!="+ ="1242ln( ) Czzx!+=+44(1 e )(1 e )4 2xxkky x+!"= +Ex 2k a constantz ax by= +d( )dzabf zx= +!d( )dyf y xx= / .Homogeneous equationsThe equation is invariant under , .. omogeneousx sx y sy!!vv' vy x y x!="=+ .Solutiond(v) vd ln constantvfxxx!==+.""1.. v' ( (v) v)iefx=!22d( ) edy xyxy y x yx!/!= +Ex 32vv2(1 v)e vdv(v v) velnv(1 v)xx!+"+!=#=.+$1 vu!+1121 1ee ( )e d e [ ]uuuu uu!!!=."vv vy x y x! !="= + .Change variablesTo evaluate integral change variables.. ln1yxyxeie x=+HomogeneousHomogeneous but for constants2 12dy x ydx x y+ +=+ +' ,'x x a y y b= += +' ' ' '.''dy dy dy dx dydx dx dx dx dx!

3 = ==' ' 2 ' 1 2' ' ' 2dy x y a bdx x y a b+ + + +=+ + + +1 2 0a b+ + =20a b+ + =3, 1ab=!=' '2 '' ' 'dy x ydx x y+=+HomogeneousThe Bernoulli equationd( ) ( ) ,1dnyP x y Q x y nx+=!To solve, change variable to 1nz y!=(1 )ndzdyn ydxdx!"=!(1 ) ( ) (1 ) ( )dzn P x z n Q xdx+!=!Gives the equationEx 42/3'y y y+ =11/3nz y y!==13'3zz!+ =x/3 Integrating factor e1/3/31xz yce!= = +/3/3/3xxze e dx!="1st order Linear1st order linearExercise:Solve the equation2y =y/x+x2/ywith initial conditiony(1) = 2. This equation is Bernoulli withn= 1. Setz=y2. Thenz z/x=x2. Integrating factorI(x) = 1/x z(x) =x[ dx x2/x+ const.] =x3/2 + const. xThusy=z1/2= x3/2 + const. x Initial conditiony(1) = 2 y(x) = x3+ 7x2 Homework1. Solve the DIFFERENTIAL equation2dydx=y(x+y)x2 homogeneous withy(1) = 1.[Answ.:y=x/(1 2 x)]2. Solve the DIFFERENTIAL equationdydx+xy=xy2 Bernoulli withy(0) = 1/2.

4 [Answ.:y= 1/(1 +ex2/2)]Exact EQUATIONS A FIRST-ORDER ODEM(x, y)dx+N(x, y)dy= 0is exact if there exists a function (x, y)such that x=M , y=N . In this case the DIFFERENTIAL equation can be recast asd =M(x, y)dx+N(x, y)dy= 0so that the solution to it is determined by (x, y) = : Solve the equationxy = 2 tany. This equation can be rewritten as2xsiny dx+x2cosy dy= 0, , M(x, y) = 2xsiny , N(x, y) =x2cosy ,which is exact because x= 2xsiny (x, y) =x2siny+ (y) y=x2cosy x2cosy+ (y) =x2cosy = constant Therefore (x, y) =x2siny+ c,and the general solution is determined byx2siny= const.: y(x) = arcsin( ) DIFFERENTIAL EQUATIONS AND FAMILIES OF CURVES General solution of a FIRST-ORDER ODEy =f(x, y)contains an arbitrary constant:y= (x, c) one curve inx, yplane for each value ofc general solution can be thought of as one-parameter family of curvesExample:y = equation y dy= x dx y2/2 = x2/2 + ,x2+y2= constant: family of circles centered at trajectories Given the family of curves representing solutions of ODEy =f(x, y),orthogonal trajectories are given by a second family of curves which aresolutions ofy = 1/f(x, y).

5 Then each curve in either family is perpendicularto every curve in the other :Find the orthogonal trajectories to the family of circlesy = x/y. Solvey =y/x . dyy= dxx lny= lnx+ ,y=cx: family of straight lines through the originHomeworka) Find the family of curves corresponding to solutions of the ODEy = (y2 x2)/(2xy).b) Find the orthogonal trajectories to the above family of curves. homogeneous equationy =f(y/x)withf(y/x) = (y/x x/y)/2solvable byy v=y/xand separation of variables x2+y2=cx: family of circles tangent toy axis at orthogonal trajectories found by solvingy = 2xy/(y2 x2) x2+y2=ky: family of circles tangent tox axis at 0 EXPLOITING FIRST-ORDER METHODS TO TREAT EQUATIONS OFHIGHER order IN SPECIAL CASES ynot present in 2nd- order equationF(x, y, y , y ) = 0 settingy =qyields 1st- order equation forq(x). xnot present in 2nd- order equationF(x, y, y , y ) = 0 settingy =q, y =dq/dx=q(dq/dy)yieldsG(y, q, dq/dy) = TsExample: homogeneous, flexible chain hanging under its own weight =linear mass densityUsing Newton s law, the shape y(x) of the chain obeys the 2nd order nonlinear DIFFERENTIAL equation y = a 1 + (y )2, a g / T Setting y = q q = a 1 + q Separation of variables 1 1 +q2dq=a dx Usingq=dy/dx= 0 atx= 0 ln(q+ 1 +q2) =ax Solving forq q=dy/dx= (eax e ax)/2 Thusy(x) =1aeax+e ax2+ constant =1acoshax+ constantThis curve is called note.

6 The problem of the catenary was the subjectof a challengeposed by Jakob Bernoulli in 1690, in response to which the problem was solvedthe following year indipendently by Johann Bernoulli, Leibniz and Find the functiony(x)obeying the DIFFERENTIAL equationy 2=x2y and the conditionsy(0) = 2,y (1) = 2.[Hint: sety =qand apply separation of variables.][Answ.:y(x) = 2(1 x) 4 ln(1 x/2)]2. Find the functiony(x)obeying the DIFFERENTIAL equationy =y eyand the conditionsy(0) = 0,y (0) = 1.[Hint:y =q;y =dq/dx=q(dq/dy); solve equation forq(y).][Answ.:y(x) = ln(1 x)]Summary No general method of solution for 1st- order ODEs beyond linear case;rather, a variety of techniques that work on a case-by-case guiding criteria: methods to bring equation to separated-variables form methods to bring equation to exact DIFFERENTIAL form transformations that linearize the equation 1st- order ODEs correspond to families of curves inx,yplane geometric interpretation of solutions EQUATIONS of higher order may be reduceable to FIRST-ORDER problems inspecial cases whenyorxvariables are missing from 2nd orderequations