2 ODEs Integrating Factors and Homogeneous Equations

Transcription

1 2 ODEs Integrating Factors an Homogeneous Equations We begin with a slightly ifferent type of equation: 2.1 Exact Equations These are ODEs whose general solution can be obtaine by simply integrating both sies of the equation. Examples Fin the general solutions of the following ODEs. a) y = 2 sin x, b) (yx2 ) = 4x 3 a) This is completely stanar we can rewrite as the integral equation integrating both sies, we obtain y = 2 cos x + c y = 2 sin(x). Then b) Here we o the same: Integrating both sies, we obtain (yx 2 ) = 4x 3 an hence yx 2 = x 4 + c. We can now ivie through by x 2 to obtain the general solution y = x 2 + c x 2. Key Point The solution of the equation is ( ) f(x)y = g(x) f(x)y = g(x) or y = 1 f(x) g(x) Terminology: In the ifferential equation b) (yx2 ) = 4x 3, the solution has two parts, since y = x 2 + c x 2. 1

2 The part x 2 is calle the efinite part. The part c (containing the arbitrary constant of integration) is x2 calle the inefinite part. If we take the efinite part of the solution, i.e. y = x 2 then (y x 2 ) = (x2 x 2 ) = (x4 ) = 4x 3. Hence y = x 2 is a solution of the original ifferential equation (b). Now consier the inefinite part of the solution, i.e. y i = c x 2, then (y i x 2 ) = ( c x 2 x2) = (c) = 0. This works generally: Useful fact: In solving an exact equation ( ) f(x)y = g(x), the solution y has two parts: y = y (x) + y i (x) where: 1. The efinite part y (x) which is a solution of the ifferential equation. 2. The inefinite part y i (x) which satisfies a simpler version of the ifferential equation in which the right-han sie is zero. Exercise: Solve the equation (y sin x) = sinx an verify that the inefinite part of the solution satisfies the equation (y sin x) = 0. Solution: Integrate both sies of the equation, y sin x = cos x + C y = cot x + Ccosec x. Hence y (x) = cot x an y i (x) = C cosec x. Now (y i(x) sin x) = (cosec x sin x) = (C) = 0. 2

3 2.1.2 Recognising exact equations The equation b) (yx2 ) = x 2 is exact but if we expan the left-han sie of this equation using the prouct rule, we have that (yx2 ) = x 2 y + y (x2 ) = x 2 y + y(2x) an so, x 2 y + 2xy = x2. (2.1) Hence the equation (2.1) is exact but it is not in stanar form Key point The equation f(x) y + yf (x) = g(x) is exact. It can be rewritten as: so that yf(x) = g(x) an y = 1 f(x) (yf(x)) = g(x) g(x). Example. Solve the equation 2x y e + 2e2x y = x 2. Note that e2x = 2e 2x. Therefore this is just the exact equation (e2x y) = x 2. Integrating both sies gives e 2x y = x3 3 + c. Thus y = x3 3 e 2x + ce 2x is the general solution. 3

4 2.2 Integrating Factors Integrating factors can be use to transform certain ODEs which are not exact into exact ODEs. As an illustration, consier x 3 y + 4x2 y = x. As you may wish to check, this is not exact. However, if we multiply through by x, we get which can be rewritten as the exact equation x 4 y + 4x3 y = x 2 (yx4 ) = x 2. Thus we can solve this as before to get yx 4 = (yx4 ) = x 2 = x3 3 + c or y = 1 3x + c x 4. The function by which we multiply a given ODE in orer to make it exact is calle the integrating factor. In the above example x is the integrating factor. This works rather generally: Funamental Algorithm Any linear ODE of the form a(x) y + b(x)y + c(x) = 0 can be transforme into an exact equation an then (hopefully!) solve. This requires the following steps: Step I: Divie by a(x) to get an equation of the form Step II: Compute the integrating factor ( I.F. = exp y + f(x)y = g(x) ) f(x) Note that we o not nee constants of integration here the IF is any function of the form e F (x) where f(x) = F (x). Step III: Multiply through by the IF to get Step IV: The LHS of this equation is simply e F (x) y = F (x) y e + f(x)ef (x) y = e F (x) g(x). (ef (x) y) an so we can solve: (ef (x) y) = e F (x) g(x). 4

5 Example: Consier the example x 3 y + 4x2 y = x from before. This can be rewritten as y + 4 x y = x 2. Thus the integrating factor (I.F.) is ( ) exp 4x 1 = exp(4 log x) = exp(log x 4 ) = x 4. Therefore, we multiply through by x 4, giving the exact ODE x 4 y + 4x3 y = x 2 ; equivalently (yx4 ) = x 4 y + 4x3 y = x 2 as before. Example: Solve the ODE y + 2y = sin x. Step 1 is alreay one! Step 2. Calculate the I.F. ( I.F. = exp ) 2 = e 2x. Step 3. Multiply through by the I.F. an write the equation in exact form. 2x y e + 2e2x y = e 2x sin x. Thus ( e 2x y ) = e 2x sin x. Step 4. Integrate both sies of the equation an fin the general solution. This is a stanar integration by parts, an I will let you check that e 2x y = e 2x y = e 2x sin x = = 1 5 e2x cos x e2x sin x + C. Example: Solve the ODE Step 1. Rewrite the ifferential equation as Step 2. Calculate the I.F. ( I.F. = exp cos x y + (sin x)y = cos2 x. ) tan(x) y + tan(x)y = cos x. = exp( ln(cos x)) = exp(ln(sec x)) = sec(x). 5

6 Step 3. Multiply through by the I.F. an write the equation in exact form. sec x y + y sec x tan x = sec(x) cos x = 1 equivalently, ( ) sec(x)y = 1 Step 4. Integrate both sies of the equation an fin the general solution: ( sec(x)y ) = which has solution sec(x)y = x + c or y = (x + c) cos(x). An electrical example: Suppose we are given an electrical circuit with a resistor an an inuctance coil an a battery. R L + E - A basic fact from physics is that, if we close the circuit we get a current i flowing clockwise aroun the circuit, with voltage rop across the resistor of ir an across the coil of L i t. Thus, E = ir + L i t. [Note: These are not formulae you nee to remember for this course.] Let s put in some numbers; say E = 10, R = 8 an L = 2; with initial conitions of i = 0 at t = 0. Thus 10 = 8i + 2 i t, or i + 4i = 5. t 6

7 This can be solve either with an IF (which equals e 4t ) or by separating variables an using (1.3.1) from page 6 of the Separation of Variables notes. Either way you fin that i = 5/4 + Ce 4t. The initial conitions give 0 = 5/4 + C an so C = 5/4 an i = e 4t. This has the graph with asymptote at i = 5/4 like those on page 7 of the Separation of Variables section. Now suppose the battery goes flat an we use the mains in place, say with voltage 2 cos(t). This means that we have to solve the equation 2 i t + 8i = 2 cos(t) or i + 4i = cos(t). t This oes now nee an IF = exp( 4t) = e 4t. Thus ( e 4t i ) 4t i = e t t + 4e4t i = e 4t cos(t). Therefore, e 4t i = e 4t cos(t) = 4 17 e4t cos(t) e4t sin(t) + C, where the integral on the RHS was solve by integration by parts (twice). Solving for i we get i = 4 17 cos(t) sin(t) + Ce 4t. Finally, solving for C from i(0) = 0 gives C = 4 17 an so, finally, i = 4 17 cos(t) sin(t) 4 17 e 4t. Example: Even quite innocent-looking equations can lea to impossible integrals. For example the equation y + xy = 1 leas to the IF = exp( x) = e x2 /2 an hence to the equation e x2 /2 y = e x2 /2, which has no close form. (In fact, you can always solve this sort of equation by writing e x2 /2 as a power series in x, but that is another story an another course.) 7

8 3 Homogenous Equations These are ifferential equations of the form: y = f(x, y) where the function f(, ) satisfies f(λx, λy) = f(x, y), for any λ. Remark: Be careful about looking up homogeneous equations in the literature since this has more than one meaning. Examples of homogeneous equations: a) f(x, y) = 1 + y x ty since f(tx, ty) = 1 + tx = 1 + y x = f(x, y). b) f(x, y) = e x y since f(tx, ty) = e tx ty = e x y = f(x, y). c) f(x, y) = x y + y x tx since f(tx, ty) = ty + ty tx = x y + y x = f(x, y). 3.1 Basic Technique for Solving Homogeneous Equations. To solve homogeneous equations, we o a substitution: z = y x (or y = zx) an use the rule y = z (zx) = x + z. We illustrate how this is one with an example. Example: Solve the ODE x y = x + y. Step 1. Divie through by x to get a homogenous equation: y = 1 + y x. Step 2. Substitute y = zx to get (xz) = 1 + z. The Key Step 3. Expan the LHS of the equation using the prouct rule, x z + z z (x) = 1 + z or x + z = 1 + z. 8

9 Step 4. Hopefully (!) this can now be solve by one of the earlier techniques. In this case we can cancel the z s to get x z = 1. This can then be solve by separating variables: This has solution z = ln x + C. z = 1 x an so z = x. Step 5. Finally substitute back z = y/x; thus y = ln(x) + C or y = x ln(x) + Cx. x Lets o this with some more examples. Example. Solve the ODE x y = y + xey/x subject to y = 1 at x = 1. Solution: Divie by x to get This is homogeneous, so we set z = y x y = y x + ey/x. or y = zx to get x z + z = (zx) = z + e z. Cancel the z to get x z = ez. This can now be solve by separating variables: e z z = x 1, with solution e z = ln x + C. Substituting back z = y/x we get e y/x = ln x + C. Plugging in y = 1 at x = 1 gives e 1 = 0 + C an so C = e 1. Thus, finally, e y/x = ln x e 1. If you like you can solve for y by taking logs, but the answer is still not very elegant. 9

10 Example. Solve the ODE y (y + x)y = x(x y) 2 Solution: This is not obviously homogeneous, but we make it so by iviing the top an bottom of the fraction on RHS by x 2 : Substitute in y = zx ( y y = x + 1) y ( ) x 1 y 2 x x z + 1)z + z = (z (1 z) 2 x z = ( z 2 + z 1 z ) z 2 We can now simplify the RHS: x z = z2 z z(1 z) 2(1 z) 1 z = z2 z z + z z 1 z = 2 1 z = 2 z 1. Thus we can separate variables: ( z 1 Integrate both sies of the equation to get 2 ) z = 1 ( z x or 2 1 ) z = 1 2 x z 2 4 z 2 = ln x + c Finally we substitute back z = y x to get y 2 4x 2 y = ln x + c. 2x With a bit of an effort you can write this as y = g(x) for some function g(x) but I really o not think it is worth the bother. 10