Transcription of ODE Cheat Sheet Nonhomogeneous Problems Series Solutions
1 ODE Cheat SheetFirst Order EquationsSeparabley (x) =f(x)g(y) dyg(y)= f(x)dx+CLinear First Ordery (x) +p(x)y(x) =f(x) (x) = exp xp( )d Integrating factor.( y) =f Exact :y(x) =1 (x)( f( ) ( )d +C)Exact0 =M(x,y)dx+N(x,y)dySolution:u(x,y) = const wheredu= u xdx+ u ydy u x=M(x,y), u y=N(x,y)Condition:My=NxNon-Exact Form (x,y) (M(x,y)dx+N(x,y)dy) =du(x,y)My=NxN x M y= ( M y N x).Special casesIfMy NxM=h(y),then (y) = exp h(y)dyIfMy NxN= h(x),then (y) = exp h(x)dxSecond Order EquationsLineara(x)y (x) +b(x)y (x) +c(x)y(x) =f(x)y(x) =yh(x) +yp(x)yh(x) =c1y1(x) +c2y2(x)Constant Coefficientsay (x) +by (x) +cy(x) =f(x)y(x) =erx ar2+br+c= 0 CasesDistinct, real roots:r=r1,2, yh(x) =c1er1x+c2er2xOne real root:yh(x) = (c1+c2x)erxComplex roots:r= i , yh(x) = (c1cos x+c2sin x)e xCauchy-Euler Equationsax2y (x) +bxy (x) +cy(x) =f(x)y(x) =xr ar(r 1) +br+c= 0 CasesDistinct, real roots:r=r1,2, yh(x) =c1xr1+c2xr2 One real root:yh(x) = (c1+c2ln|x|)xrComplex roots.
2 R= i ,yh(x) = (c1cos( ln|x|) +c2sin( ln|x|))x Nonhomogeneous ProblemsMethod of Undetermined Coefficientsf(x)yp(x)anxn+ +a1x+a0 Anxn+ +A1x+A0aebxAebxacos x+bsin xAcos x+Bsin xModified Method of Undetermined Coefficients:if anyterm in the guessyp(x) is a solution of the homogeneousequation, then multiply the guess byxk, wherekis thesmallest positive integer such that no term inxkyp(x) is asolution of the homogeneous of OrderHomogeneous CaseGiveny1(x) satisfiesL[y] = 0,find second linearly independentsolution asv(x) =v(x)y1(x). z=v satisfies a separable CaseGiveny1(x) satisfiesL[y] = 0,find solution ofL[y] =fasv(x) =v(x)y1(x). z=v satisfies a first order linear of Variation of Parametersyp(x) =c1(x)y1(x) +c2(x)y2(x)c 1(x)y1(x) +c 2(x)y2(x) = 0c 1(x)y 1(x) +c 2(x)y 2(x) =f(x)a(x)ApplicationsFree Fallx (t) = gv (t) = g+f(v)Population DynamicsP (t) =kP(t)P (t) =kP(t) bP2(t)Newton s Law of CoolingT (t) = k(T(t) Ta)Oscillationsmx (t) +kx(t) = 0mx (t) +bx (t) +kx(t) = 0mx (t) +bx (t) +kx(t) =F(t)Types of Damped OscillationOverdamped,b2>4mkCritically Damped,b2= 4mkUnderdamped,b2<4mkNumerical MethodsEuler s Methody0=y(x0),yn=yn 1+ xf(xn 1,yn 1), n= 1.
3 , SolutionsTaylor Methodf(x) n=0cnxn, cn=f(n)(0)n!1. Differentiate DE Apply initial Find Taylor Insert coefficients into Series form fory(x).Power Series Solution1. Lety(x) = n=0cn(x a) Findy (x), y (x).3. Insert expansions in Collect like terms using Find recurrence Solve for coefficients and insert iny(x) and Singular Pointsy +a(x)y +b(x)y= 0. x0is aOrdinary point:a(x),b(x) real analytic in|x x0|< RRegular singular point: (x x0)a(x), (x x0)2b(x) haveconvergent Taylor Series aboutx= singular point: Not ordinary or regular Method1. Lety(x) = n=0cn(x x0)n+ Obtain indicial equationr(r 1) +a0r+ Find recurrence relation based on types of roots ofindicial Solve for coefficients and insert iny(x) TransformsTransform Pairsccseat1s a,s > atnn!
4 Sn+1, s >0sin t s2+ 2cos tss2+ 2sinhatas2 a2coshatss2 a2H(t a)e ass, s >0 (t a)e as, a 0,s >0 Laplace Transform PropertiesL[af(t) +bg(t)] =aF(s) +bG(s)L[tf(t)] = ddsF(s)L[dfdt]=sF(s) f(0)L[d2fdt2]=s2F(s) sf(0) f (0)L[eatf(t)] =F(s a)L[H(t a)f(t a)] =e asF(s)L[(f g)(t)] =L[ t0f(t u)g(u)du] =F(s)G(s)Solve Initial Value Problem1. Transform DE using initial Solve forY(s).3. Use transform pairs, partial fraction decomposition, toobtainy(t).Special FunctionsLegendre PolynomialsPn(x) =12nn!dndxn(x2 1)n(1 x2)y 2xy +n(n+ 1)y= 0.(n+ 1)Pn+1(x) = (2n+ 1)xPn(x) nPn 1(x), n= 1,2,..g(x,t) =1 1 2xt+t2= n=0Pn(x)tn,|x| 1,|t|< Functions,Jp(x), Np(x)x2y +xy + (x2 p2)y= Functions (x) = 0tx 1e tdt, x >0.
5 (x+ 1) =x (x).Systems of Differential EquationsPlanar Systemsx =ax+byy =cx+ (a+d)x + (ad bc)x= Formx =(x y )=(a bc d)(xy) t Av= ProblemAv= Eigenvalues: det(A I) = 0 Find Eigenvectors (A I)v= 0 for each .CasesReal, Distinct Eigenvalues:x(t) =c1e 1tv1+c2e 2tv2 Repeated Eigenvalue:x(t) =c1e tv1+c2e t(v2+tv1),whereAv2 v2= Conjugate Eigenvalues:x(t) =c1Re(e t(cos t+isin t)v) +c2Im(e t(cos t+isin t)v).Solution BehaviorStable Node: 1, 2< Node: 1, 2> : 1 2< : =i .Stable Focus: = +i , < Focus: = +i , > SolutionsLetx = eigenvalues iFind eigenvectorsvi=(vi1vi2)Form the Fundamental Matrix Solution: =(v11e 1tv21e 2tv12e 1tv22e 2t)General Solution:x(t) = (t)CforCFindC:x0= (t0)C C= 1(t0)x0 Particular Solution:x(t) = (t) 1(t0) Matrix solution: (t) = (t) 1(t0).
6 Particular Solution:x(t) = (t) : =A , (t0) = Matrix Solutionsx(t) = (t)C+ (t) tt0 1(s)f(s)dsx(t) = (t)x0+ (t) tt0 1(s)f(s)ds2 2 Matrix Inverse(a bc d) 1=1detA(d b c a)
