Ordinary Differential Equations - IIT Guwahati

Ordinary Differential Equations(MA102 Mathematics II)Shyamashree UpadhyayIIT GuwahatiShyamashree Upadhyay ( IIT Guwahati )Ordinary Differential Equations1 / 25

First order ODE sWe will now discuss different methods of solutions of first order ODEs. The first type ofsuch ODEs that we will consider is the following:DefinitionSeparable variables: A first order differential equation of the formdy g(x)h(y)dxis called separable or to have separable variables.Such ODEs can be solved by direct integration:dydyWrite dx g(x)h(y) as h(y) g(x)dx and then integrate both sides!Exampledye x dx e y e 2x yThis equation can be rewritten as dx e x e y e 3x y , which is the same asdy e y (e x e 3x ). This equation is now in separable variable form.dxShyamashree Upadhyay ( IIT Guwahati )dyOrdinary Differential Equations2 / 25

Losing a solution while separating variablesSome care should be exercised in separating the variables, since the variable divisorscould be zero at certain points. Specifically, if r is a zero of the function h(y), thendysubstituting y r in the ODE dx g(x)h(y) makes both sides of the equation zero; in otherdywords, y r is a constant solution of the ODE dx g(x)h(y). But after variables aredyseparated, the left hand side of the equation h(y) g(x)dx becomes undefined at r. As aconsequence, y r might not show up in the family of solutions that is obtained afterdyintegrating the equation h(y) g(x)dx.Recall that such solutions are called Singular solutions of the given ODE.ExampleObserve that the constant solution y 0 is lost while solving the IVPseparable variables method.Shyamashree Upadhyay ( IIT Guwahati )Ordinary Differential Equationsdydx xy; y(0) 0 by3 / 25

First order linear ODEsRecall that a first order linear ODE has the forma1 (x)dy a0 (x)y g(x).dx(1)DefinitionA first order linear ODE (of the above form (1)) is called homogeneous if g(x) 0 andnon-homogeneous otherwise.DefinitionBy dividing both sides of equation (1) by the leading coefficient a1 (x), we obtain a moreuseful form of the above first order linear ODE, called the standard form, given bydy P(x)y f (x).dx(2)Equation (2) is called the standard form of a first order linear ODE.Shyamashree Upadhyay ( IIT Guwahati )Ordinary Differential Equations4 / 25

TheoremTheoremExistence and Uniqueness: Suppose a1 (x), a0 (x), g(x) C((a, b)) and a1 (x) , 0 on (a, b)and x0 (a, b). Then for any y0 R, there exists a unique solution y(x) C 1 ((a, b)) to theIVPa1 (x)Shyamashree Upadhyay ( IIT Guwahati )dy a0 (x)y g(x); y(x0 ) y0 .dxOrdinary Differential Equations5 / 25

Solving a first order linear ODESteps for solving a first order linear ODE:(1) Transform the given first order linear ODE into a first order linear ODE in standard formdy P(x)y f (x).dxR(2) Multiply both sides of the equation (in the standard form) by e P(x)dx . Then the resultingequation becomesRd[yedxP(x)dxR] f (x)eP(x)dx(3).(3) Integrate both sides of equation (3) to get the solution.ExampleSolve x dx 4y x6 e x .The standrad form of this ODE isdyR 4dxequation by exdydx ( 4)y x5 e x . Then multiply both sides of thisxand integrate.Shyamashree Upadhyay ( IIT Guwahati )Ordinary Differential Equations6 / 25

Differential of a function of 2 variablesDefinitionDifferential of a function of 2 variables: If f (x, y) is a function of two variables withcontinuous first partial derivatives in a region R of the xy-plane, then its differential d f isdf f fdx dy. x yIn the special case when f (x, y) c, where c is a constant, we haveTherefore, we have d f 0, or in other words, f x 0 and f y 0. f fdx dy 0. x ySo given a one-parameter family of functions f (x, y) c, we can generate a first orderODE by computing the differential on both sides of the equation f (x, y) c.Shyamashree Upadhyay ( IIT Guwahati )Ordinary Differential Equations7 / 25

Exact differential equationDefinitionA differential expression M(x, y)dx N(x, y)dy is an exact differential in a region R of thexy-plane if it corresponds to the differential of some function f (x, y) defined on R. A firstorder differential equation of the formM(x, y)dx N(x, y)dy 0is called an exact equation if the expression on the left hand side is an exact differential.Example: 1) x2 y3 dx x3 y2 dy 0 is an exact equation since x2 y3 dx x3 y2 dy d(2) ydx xdy 0 is an exact equation since ydx xdy d(xy).3)ydx xdyy2 0 is an exact equation sinceShyamashree Upadhyay ( IIT Guwahati )ydx xdyy2x3 y3).3 d( yx ).Ordinary Differential Equations8 / 25

Criterion for an exact differentialTheoremLet M(x, y) and N(x, y) be continuous and have continuous first partial derivatives in arectangular region R defined by a x b, c y d. Then a necessary and sufficientcondition for M(x, y)dx N(x, y)dy to be an exact differential is M N. y xExampleSolve the ODE (3x2 4xy)dx (2x2 2y)dy 0.This equation can be expressed as M(x, y)dx N(x, y)dy 0 where M(x, y) 3x2 4xy andN(x, y) 2x2 2y. It is easy to verify that M N 4x. Hence the given ODE is exact. y xWe have to find a function f such that f x f x M 3x2 4xy and f y N 2x2 2y. Now 3x 4xy f (x, y) (3x 4xy)dx x 2x y φ(y) for some function φ(y) of y. f 2x2 2y and f (x, y) x3 2x2 y φ(y) together imply that y202x φ (y) 2x2 2y φ(y) y2 c1 for some constant c1 . Hence the solution isf (x, y) c or x3 2x2 y y2 c1 c.2R232AgainShyamashree Upadhyay ( IIT Guwahati )Ordinary Differential Equations9 / 25

Converting a first order non-exact DE to exact DEConsider the following example:ExampleThe first order DE ydx xdy 0 is clearly not exact. But observe that if we multiply bothsides of this DE by y12 , the resulting ODE becomes dx yx2 dy 0 which is exact!yDefinitionIt is sometimes possible that even though the original first order DEM(x, y)dx N(x, y)dy 0 is not exact, but we can multiply both sides of this DE by somefunction (say, µ(x, y)) so that the resulting DE µ(x, y)M(x, y)dx µ(x, y)N(x, y)dy 0becomes exact. Such a function/factor µ(x, y)is known as an integrating factor for theoriginal DE M(x, y)dx N(x, y)dy 0.Remark: It is possible that we LOSE or GAIN solutions while multiplying a ODE by anintegrating factor.Shyamashree Upadhyay ( IIT Guwahati )Ordinary Differential Equations10 / 25

How to find an integrating factor?We will now list down some rules for finding integrating factors, but before that, we needthe following definition:DefinitionA function f (x, y) is said to be homogeneous of degree n if f (tx, ty) tn f (x, y) for all (x, y)and for all t R.Example1) f (x, y) x2 y2 is homogeneous of degree 2.y2) f (x, y) tan 1 ( x ) is homogeneous of degree 0.3) f (x, y) x(x2 y2 )y22is homogeneous of degree 1.4) f (x, y) x xy 1 is NOT homogeneous.Shyamashree Upadhyay ( IIT Guwahati )Ordinary Differential Equations11 / 25

How to find an integrating factor? contd.DefinitionA first order DE of the formM(x, y)dx N(x, y)dy 0is said to be homogeneous if both M(x, y) and N(x, y) are homogeneous functions of thesame degree.NOTE: Here the word “homogeneous does not mean the same as it did for first orderlinear equation a1 (x)y0 a0 (x)y g(x) when g(x) 0.Some rules for finding an integrating factor: Consider the DEM(x, y)dx N(x, y)dy 0.Rule 1: If ( ) is a homogeneous DE with M(x, y)x N(x, y)y , 0, then( )1Mx Nyis anintegrating factor for ( ).Shyamashree Upadhyay ( IIT Guwahati )Ordinary Differential Equations12 / 25

How to find an integrating factor? contd.Rule 2: If M(x, y) f1 (xy)y and N(x, y) f2 (xy)x and Mx Ny , 0, where f1 and f2 are1functions of the product xy, then Mx Nyis an integrating factor for ( ).Rule 3: IfRule 4: If M N y xN M N y xMR f (x) (function of x-alone), then ef (x)dx F(y) (function of y-alone), then e RShyamashree Upadhyay ( IIT Guwahati )is an integrating factor for ( ).F(y)dyOrdinary Differential Equationsis an integrating factor for ( ).13 / 25

Proof of Rule 3Proof.Let f (x) M N y xR. To show: µ(x) : e (µM) (µN). y xf (x)dxNis an integrating factor. That is, to show Since µ is a function of x alone, we have y(µM) µ M. Also ySo we must have: N] µ0 (x)N , or equivalently we must have,µ(x)[ M y xµ0 (x)µ(x) (µN) x µ0 (x)N µ(x) N. x f (x),Rwhich is anyways true since µ(x) : ef (x)dx. The proof of Rule 4 is similar. The proof of Rule 2 is an exercise.Shyamashree Upadhyay ( IIT Guwahati )Ordinary Differential Equations14 / 25

Another rule for finding an I.F.Shyamashree Upadhyay ( IIT Guwahati )Ordinary Differential Equations15 / 25

Solution by substitutionOften the first step of solving a differential equation consists of transforming it into anotherdifferential equation by means of a substitution.dyFor example, suppose we wish to transform the first order differential equation dx f (x, y)by the substitution y g(x, u), where u is regarded as a function of the variable x. If gpossesses first partial derivatives, then the chain ruledy g dx g du dx x dx u dxdy g x (x, u) gu (x, u) du. The original differential equation dx f (x, y) now becomesdxdudug x (x, u) gu (x, u) dx f (x, g(x, u)). This equation is of the form dx F(x, u), for somefunction F . If we can determine a solution u φ(x) of this last equation, then a solution ofthe original differential equation will be y g(x, φ(x)).givesdydxShyamashree Upadhyay ( IIT Guwahati )Ordinary Differential Equations16 / 25

Use of substitution : Homogeneous equationsRecall: A first order differential equation of the form M(x, y)dx N(x, y)dy 0 is said to behomogeneous if both M and N are homogeneous functions of the same degree.Such equations can be solved by the substitution : y vx.ExampleSolve x2 ydx (x3 y3 )dy 0.Solution: The given differential equation can be rewritten asdydx x2 y.x3 y3Let y vx, thendvthis in the above equation, we get v x dx in other words,is now in separable variables form.dydv v x dx. Puttingdx31 vdx( v4 )dv x , whichShyamashree Upadhyay ( IIT Guwahati )Ordinary Differential Equationsv.1 v3Or17 / 25

DE reducible to homogeneous DEFor solving differential equation of the formdyax by c dx a0 x b0 y c0use the substitutionx X h and y Y k, ifz ax by, ifaa0 aa0,b,b0where h and k are constants to be determined.b.b0ExampleSolvedydx x y 4.3x 3y 5Observe that this DE is of the formdydx ax by cwhere aa0 bb0 .a0 x b0 y c0dydzhave dx 1 dx. PuttingUse the substitution z x y. Then wethese in the given DE, wedzz 4get dx 1 3z 5, or in other words, 3z 5dz dx. This equation is now in separable4z 9variables form.Shyamashree Upadhyay ( IIT Guwahati )Ordinary Differential Equations18 / 25

DE reducible to homogeneous DE, contd.ExampleSolvedydx x y 4.x y 6Observe that this DE is of the form dx a0 x b0 y c0 where 1 aa0 , bb0 1.Put x X h and y Y k, where h and k are constants to be determined. Then we havedx dX , dy dY anddyax by cX Y (h k 4)dY .dXX Y (h k 6)( )If h and k are such that h k 4 0 and h k 6 0, then ( ) becomesdYX Y dXX Ywhich is a homogeneous DE. We can easily solve the systemh k 4h k 6of linear equations to detremine the constants h and k!Shyamashree Upadhyay ( IIT Guwahati )Ordinary Differential Equations19 / 25

Reduction to separable variables formA differential equation of the formdy f (Ax By C),dxwhere A, B, C are real constants with B , 0 can always be reduced to a differentialequation with separable variables by means of the substitution u Ax By C .Observe that since B , 0, we getimplies thatdydxuB AxB y CB , or in other words, y B1 ( du) AB . Hence we have B1 ( du) dxdxother words, we haveduA B f (u)Shyamashree Upadhyay ( IIT Guwahati )AB f (u), that is,dudxuB AB x CB . This A B f (u). Or in dx, which is now in separable variables form.Ordinary Differential Equations20 / 25