ORDINARY DIFFERENTIAL EQUATIONS

ORDINARY DIFFERENTIAL EQUATIONS Basic concepts: Find y(x) where x is the independent and y the dependent varible, based on an equation involving x, y(x), y 0 (x),...e.g.: y 00 (x) = 1+y(x) y0 (x) 1+x or, equivalently y 00 = 1+yy0 1+x The highest derivative (second) is the order of the equation. Solution is normally a family of functions, with as many extra parameters (C 1, C ) as the order of the equation. We will first stu first-order ODE, then higher-order ODE, but, almost exclusively linear (in y and its derivatives) with constant coefficients, e.g: y 00 y 0 +3y = e x. WhentheRHSofiszero,theequationiscalledhomogenous. A set (system) of ODE has several dependent (unknown) functions y 1,y,y 3,... of a single independent variable x. We will stu only the first order, linear set of ODE with constant coefficients (matrix algebra). Partial differential equations have a single dependent variable and several independent variables (partial derivatives). We may not have time to discuss these. General form: First-Order Differential Equations y 0 = f(x, y) (visualize graphically). Analytically, this ODE can be solved only in a handful of special cases. The family of solutions usually (but not always) covers the whole x y plane by curves which don t intersect (one solution passes through each point). This means: given y(x 0 )=y 0 (initial condition), there is a unique solution. Let s go over the special cases now. Trivial equation:

y 0 = f(x) General solution (if we can integrate f) Z y(x) = f(x)dx + C EXAMPLE: y 0 =sin(x). Solution: y(x) = cos(x) +C. Separable equation y 0 = h(x) g(y) Solution: Writing y 0 as, we can separate x and y: dx g(y) = h(x)dx and integrate each side individually (w/r to y and x, respectively - don t forget to add C). Ifwecanthen solve for y, we have an explicit solution, if not, we leave it in the implicit form. EXAMPLES: 1. y 0 = x y y = xdx ln y = x + C y = ±e C e x Ce x

. 9yy 0 +4x = 0 9y = 4xdx 9 y y + 4 9 x = C = 4 x + C family of ellipses centered on the origin, with the vertical versus horizontal diameter in the :3 ratio. 3. y 0 = xy y = xdx ln y = x + C y = Ce x 4. (1 + x )y 0 +1+y =0 with y(0) = 1 (initial value problem). Solution: = dx 1+y 1+x arctan(y) = arctan(x)+ C arctan(c) arctan(x) y = tan (arctan(c) arctan(x)) = C x 1+Cx To find C we solve 1= Answer: y(x) = 1 x Check: (1 + x ) d dx 1+x. 1 x 1+x C 0 C =1. 1+C 0 +1+ 1 x 1+x =0. 3

Scale-independent equation ³ y y 0 = g x (RHS invariant under x ax and y ay). Solve by introducing a new dependent variable u(x) = y(x) x y 0 = u + xu 0. Substitute into original equation: which is separable in x and u: xu 0 = g(u) u du g(u) u = dx x Solve as such, and then go back to y(x). EXAMPLES: 1. xyy 0 y + x = 0 y 0 = y x x y xu 0 = u +1 u udu u +1 = dx x ln(1 + u ) = ln x + C u +1 = C x y + x Cx = 0 y +(x C) = C y(x) =x u(x) Family of circles having a center at any point of the x-axis, and being tangent to the y-axis 4

. x y 0 = y + xy + x ³ y y 0 y = + x x +1 xu 0 = u +1 du 1+u = dx x arctan(u) = ln x + C u = tan(ln x + C) y = x tan(ln x + C) Modified Scale-Independent y 0 = y ³ y x + g h(x) x The same substitution yields which is also separable. EXAMPLE: xu 0 = g(u) h(x) y 0 = y x + x3 cos(x ) y xu 0 = x cos(x ) u udu = x cos(x ) dx u = sin(x )+ C u = ± p sin(x )+C y = ±x p sin(x )+C Any Other Smart Substitution (usually suggested), which makes the equation separable. 5

EXAMPLES: 1. (x 4y +5)y 0 + x y +3=0 Suggestion: introduce: v = x y, i.e. y = x v and y 0 = 1 v0 (v +5) 1 v0 + v +3 = 0 (v + 5 )v0 +v + 11 = 0 v + 5 dv = dx µ 1 v + 11 4 1 4 v + 11 4 dv = dx v 1 4 ln v + 11 4 = x + C x y 1 4 ln x y + 11 4 = x + C. y 0 cos y + x sin y =x seems to suggest v =siny, sincev 0 = y 0 cos y. The new equation is thus simply v 0 + xv =x which is separable and can be solved as such: dv v = xdx ln v = x +lnc v = Ce x v = +Ce x ³ Finally, y =arcsinv =arcsin +Ce x. 6

Linear equation y 0 + g(x) y = r(x) The solution is constructed in two stages: 1. Solve the homogeneous part y 0 = g(x) y, which is separable, thus: y h (x) =c e R g(x)dx. Assume c to be a function of x, substitute c(x) e R g(x)dx back into the full equation, and solve the resulting trivial differential equation for c(x). EXAMPLES: 1. First solve y 0 + y x = sin x x y 0 + y x = 0 y = dx x ln y = ln x +lnc y = c x Now substitute this to the original equation: c 0 x c x + c = sin x x x c 0 = sinx c(x) = cos x + C y(x) = cos x x + C x Note that the solution has always the form of y p (x)+cy h (x), where y p (x) is a particular solution to the full equation, and y h (x) solves the homogeneous equation only. 7

Let us verify the former: d dx ³ cos x x cos x x = sin x x. y 0 y = e x First y 0 y = 0 y = dx y = ce x Substitute: c 0 e x + ce x ce x = e x c 0 = e x c(x) = e x + C y(x) = e x + Ce x 3. xy 0 + y +4=0 Homogeneous part: y = dx x ln y = ln x +lnc y = c x Substitute: c 0 c x + c x = 4 c(x) = 4x + C y(x) = 4+ C x 8

4. with y(0) = 1. Homogeneous part: y 0 + y tan(x) =sin(x) sin xdx = y cos x ln y = ln cos x +lnc y = c cos x Substitute: c 0 cos x c sin x + c sin x = sinx cos x c 0 = sinx c(x) = cosx + C y(x) = cos x + C cos x To find the value of C, solve: The final answer is thus: To verify: 1= +C C =3 y(x) = cos x +3cosx d cos x +3cosx + dx cos x +3cosx sin x =cosxsin x cos x 5. with y(1) = 0. x y 0 +xy x +1=0 9

Homogeneous part: y = dx x ln y = ln x + lnc y = c x Substitute: c 0 c x + c x x +1 = 0 c 0 = x 1 c = x x + C y = 1 1 x + C x To meet the initial-value condition: 0= 1 1+C C = 1 Final answer: Verify: x d dx y = (1 x) x µ µ (1 x) (1 x) +x x +1 0. x x 6. y 0 y x = x cos(3x) First: y = dx x ln y = ln x +lnc y = cx 10

Substitute: c 0 x +cx cx = x cos(3x) c 0 = cos(3x) c = sin(3x) + C 3 y = x 3 sin(3x)+cx To verify the particular solution: µ d x dx 3 sin(3x) x 3 sin(3x) =x cos(3x) Bernoulli equation where a is a specific, constant exponent. y 0 + f(x) y = r(x) y a Introducing a new dependent variable u = y 1 a, i.e. y = u 1 1 a, one gets: Multiplying by (1 a)u a 1 a 1 1 a u 1 1 a 1 u 0 + f(x) u 1 1 a results in: = r(x) u a 1 a u 0 +(1 a)f(x) u =(1 a)r(x) which is linear in u 0 and u. Theansweristheneasilyconvertedbacktoy = u 1 1 a. EXAMPLES: 1. y 0 + xy = x y Bernoulli, a = 1, f(x) x, r(x) x, implying u 0 +xu =x 11

where y = u 1. Solving as linear: du u = xdx ln u = x +lnc u = c e x Substitute:. c 0 e x xce x +xce x = x c 0 = xe x c(x) = e x + C u(x) = 1+Ce x p y(x) = ± 1+Ce x (one can easily check that this is a solution with either the + or the sign). xy 0 =10x 3 y 5 + y (terms reshuffled a bit). Bernoulli with a =5, f(x) = 1, and r(x) =5x x This implies u 0 + x u = 0x with y = u 1 4. Solving as linear: du u = dx x ln u = ln x + lnc u = c x 1

Substituted back into the full equation: c 0 x c x 3 + c x 3 = 0x c 0 = 0x 4 c(x) = 4x 5 + C u(x) = 4x 3 + C x y(x) = ± µ 4x 3 + Cx 1 4. 3. xyy 0 +(x 1)y = x e x Bernoulli with a = 1, f(x) = x 1, and r(x) = x x ex This translates to: u 0 + x 1 x u = xex with y = u 1. Solving homogeneous part: du u = ( 1 1) dx x ln u = ln x x +lnc u = cxe x Substituted: c 0 xe x + ce x cxe x +(x 1)ce x = xe x c 0 = e x c(x) = 1 ex + C u(x) = x ex + Cxe x r x y(x) = ± ex + Cxe x 13

Exact equation General idea: Suppose we have a function of x and y, f(x, y) say. Then f f dx + x y is its total differential, representing the function s increase from (x, y) to (x + dx, y + ). By making this equal to zero (a differential equation, called exact), we are effectively saying that f(x, y) =C (and this is its implicit solution). EXAMPLE: Suppose This means that has a simple solution f(x, y) =x y x (xy ) dx + x =0 x y x = C y = x + C x Note that the differential equation can be also written as: (linear). y 0 = 1 xy x We must now try to reverse the process, i.e. given a differential equation, find f(x, y). There are then two issues to be settled: 1. Howdoweknowthatanequationis exact?. Knowing it is, how do we solve it? 14

To answer the first question, we recall that f x y f y x Thus, g(x, y) dx + h(x, y) =0is exact if and only if g y h x As to solving the equation, we proceed in three stages: 1. Find Z G(x, y) = g(x, y) dx (considering y a constant).. Construct H(y) =h(x, y) G y [must be a function of y only, as H = h G = g g 0]. x x x y y y 3. Z f(x, y) =G(x, y)+ H(y) Proof: f = G f = g and = G + H = h. x x y y EXAMPLE: x sin(3y) dx + 3x cos(3y)+y =0 Let us firstverifythattheequationis exact: x sin(3y) =6xcos 3y y 3x cos(3y)+y =6xcos 3y x Solving it: G = x sin(3y) H = 3x cos(3y)+y 3x cos(3y) =y f(x, y) = x sin(3y)+y Answer: y + x sin(3y) =C (implicit form). 15

Integrating Factors Any first-order ODE (e.g. y 0 = y ) canbeexpandedtomakeitlook like an x exact equation: dx = y x ydx x = 0 But since (y) (x) 6=, this equation is not exact. y x The good news is that, theoretically, there is always a function of x and y, say F (x, y), which can multiply the equation to make it exact. This function is called an integrating factor. The bad news is that there is no general procedure for finding F (x, y). Yet, there are two special cases when it is possible: Let us write the differential equation in its look-like-exact form of where P y P (x, y)dx + Q(x, y) =0 Q 6=.Onecanfind an integrating factor from x 1. d ln F dx P = Q y x Q iff the right hand side of this equation is a function of x only Proof: FPdx+ FQ =0is exact when (FP) y = (FQ) x which is the same as F P y = df dx Q + F Q x assuming that F is a function of x only. Solving for df dx F results in P y Q x. Q When the last expression contains no y, we simply integrate it (with respect to x) to find ln F. 16

. or from d ln F = Q x P y P iff the right hand side is a function of y only. EXAMPLES: 1. Let us try solving our ydx x =0. Since P Q y x Q = x we have Z dx ln F = x = lnx (noneedtobotherwithaconstant) F = 1. Thus y dx 1 =0must x x x be exact (check it). Solving it gives y = C, or y = Cx. x Note that there is infinitely many integrating factors, if F (x, y) is one, so is F (x, y) R(f(x, y)), where R is an arbitrary function.. cosy +4x dx = x sin y Since we get P Q y x Q = siny +siny x sin y ln F = Z 1 dx =lnx x F = x. = 1 x x cos y +4x 3 dx x sin y=0 is therefore exact, and can be solved as such: x cos y + x 4 = C µ C y = arccos x x 17

3. (3xe y +y) dx +(x e y + x) =0 Trying again which means that P Q y x Q = xey +1 x e y + x = 1 x ln F = Z dx x =lnx F = x. is exact. Solving it yields: Clairaut equation: (3x e y +xy) dx +(x 3 e y + x ) =0 x 3 e y + x y = C y = xy 0 + g(y 0 ) where g is an arbitrary function. The idea is to introduce p y 0,differentiate the original equation with respect to x, obtaining p = p + xp 0 + p 0 g 0 (p) p 0 (x + g 0 (p)) = 0. This implies that either p y 0 = C y = xc + g(c) which represents a family of regular solutions (all straight lines), or x = g 0 (p) which, when solved for p and substituted back into y = xp + g(p) provides the so called singular solution (an envelope of the regular family). EXAMPLE: (y 0 ) xy 0 + y =0 18

(terms reshuffled a bit) is solved by either or y = Cx C x = p p = x y = xp p = x 4 (singular solution). Note that for an initial condition below or at the parabola two possible solutions exist, above the parabola there is none. Final possibility: When an equation appears more complicated in terms of y rather than x, e.g. (x + y 4 )y 0 = y one can try reversing the role of x and y. Allittakesistoreplacey 0 by 1, dx dx for example (using the previous equation): dx =x y + y3 The last equation is linear and can be solved as such: dx = x y ln x = ln y + c x(y) = y c(y) 19

Substituted into the full equation: yc + y dc dc = y c y = y + y3 c(y) = y C x = y4 Cy This can now be solved for y in terms of x, togetasolutiontotheoriginal equation: y = ± p C ± C +x. Of Geometric Kind: Applications 1. Find a curve such that, from each of its points, the distance to the origin is thesameasthedistancetotheintersectionofitsnormal (i.e. perpendicular straight line) with the x-axis. Solution: Supposey(x) is the equation of the curve (yet unknown). The equation of the normal is Y y = 1 (X x) y0 where (x, y) [fixed] are the points of the curve, and (X, Y ) [variable] are the points of the normal. This normal intersects the x-axis at Y =0and X = yy 0 + x. The distance between this and the original (x, y) is p (yy0 ) + y the distance from (x, y) to (0, 0) is p x + y These two distances are equal when y (y 0 ) = x 0

or y 0 = ± x y This is a separable differential equation easy to solve: y ± x = C The curves are either circles centered on (0, 0), orhyperbolas[withy = ±x as special cases].. Find a curve whose normals (all) pass through the origin. Solution (we can guess the answer, but let us do it properly): Into the same equation of the curve s normal (see above), we substitute 0 for both X and Y, since the straight line must pass through (0, 0). This gives: which is simple to solve: y = x y 0 y = xdx x + y = C (circles centered on the origin we knew that!). 3. A family of curves covering the whole x y plane enables one to draw lines perpendicular to these curves. The collection of all such lines is yet another family of curves orthogonal (i.e. perpendicular) to the original family. If we can find the differential equation y 0 = f(x, y) having the original family of curves as its solution, we can find the corresponding orthogonal family by solving y 0 = 1. The next set of examples relates to this. f(x,y) 1. The original family is described by x +(y C) = C with C arbitrary (i.e. collection of circles tangent to the x-axis at the origin). To find the corresponding differential equation, we differentiate the original equation with respect to x: x +(y C)y 0 =0 1

, and then eliminate C by solving the original equa- solve for y 0 = x tion for C, thus: C y x + y Cy =0 C = x + y y further implying y 0 x = x +y y = xy x y y To find the orthogonal family,wesolve y 0 = y x xy [scale-independent equation solved earlier]. The answer was: (x C) + y = C i.e. collection of circles tangent to the y-axis at the origin.. Let the original family be circles centered on the origin (it should be clear what the orthogonal family is, but again, let s solve it anyhow): describes the original family, x + y = C x +yy 0 =0 is the corresponding differential equation, equivalent to y 0 = x y The orthogonal family is the solution to y 0 = y x y = dx x y = Cx (straight lines passing through the origin).

3. Let the original family be described by y = x + C (the y = x parabola slid horizontally). The corresponding differential equation is yy 0 =1 the orthogonal equation: Answer: (try to visualize the curves). 4. Finally, let us start with y 0 = y. ln y = x + C y = Ce x y = Cx (all parabolas tangent to the x-axis at the origin). Differentiating: implying (since C = y x ) The orthogonal equation is y 0 =Cx y 0 = y x y 0 = x y y + x = C (collection of ellipses centered on the origin, with the x-diameter being times bigger than the y-diameter). 4. The position of four ships on the ocean is such that the ships form vertices of a square of length L. Atthesameinstanteachshipfires a missile that directs its motion towards the missile on its right. Assuming that the four 3

missiles fly horizontally and with the same constant speed, find the path of each. Solution: Let us place the origin at the center of the original square. It should be obvious that when we find one of the four paths, the other three can be obtained just by rotating it by 90, 180 and 70 degrees. This is actually true for the missiles positions at any instant of time. Thus, if a missile is at (x, y), the one to its right is at (y, x) [(x, y) rotated by 90 o ]. If y(x) is the resulting path for the first missile Y y = y 0 (X x) is the straight line of its immediate direction. This straight line must pass through (y, x) [that s where the other missile is, at the moment]. This means that, when we substitute y and x for X and Y, respectively, the equation must hold true: x y = y 0 (y x) And this is the differential equation to solve (scale independent): xu 0 + u = 1+u 1 u xu 0 = 1+u 1 u 1 u 1+u du = dx x arctan(u) 1 ln(1 + u ) = ln x + C e arctan(u) 1+u = Cx This solution becomes a lot easier to understand in polar coordinates [θ = arctan( y x ),andr = p x + y ], whereitlookslikethis: r = eθ C (a spiral). For the upper-right-corner missile, C = L eπ/4, implying r = L exp(θ π/4) The total length travelled (till the missiles meet at r =0)is L. 4

In Physics: If a hole is made at a bottom of a container, water will flowoutattherate of a h, where a is established based on the size of the opening, but to us it is simply a constant, and h is the height of the (remaining) water, which varies in time. Time t is the independent variable. Find h(t) as a function of t for: 1. A cylindrical container of radius r and height h 0. Solution: First we have to establish the volume V of the remaining water as a function of height. In this case we get simply V (h) =πr h Differentiating with respect to t we get: dv dt = πr dh dt This in turn must be equal to a h, since the rate at which the water is flowing out must be equal to the rate at which its volume is decreasing. Thus πr h = a h where h dh. This is a simple (separable) differential equation, which we dt solve by dh = a h πr dt or equivalently h 1/ 1 = at πr + C h = at πr + p h 0 t = πr a ³p h0 h Subsidiary: Whatpercentageoftimeisspentemptyingthelast0%ofthe container? Solution: t 100 = πr p h0 a 5

is the time to fully empty the container. Ã t 80 = πr ph0 a r! h0 5 is the time it takes to empty the first 80% of the container. The answer: r t 100 t 80 1 = t 100 5 =44.7%. A conical container with the top radius (at h 0 ) equal to r. Solution: V (h) = 1 µ 3 πh r h h 0 (Notethatonefifth of the full volume corresponds to 1 5 1/3 h 0, i.e. 58.48% of the full height!). Thus πr h h = a h 3h 0 is again a separable equation: This implies and implying h 3/ dh = 3ah 0 πr dt h 5/ = 15ah 0 4πr t + h5/ 0 t = 4πr 15ah 0 t 100 = 4πr h 0 15a " 1 t 80 = 4πr h 0 15a t 1 t 0.8 t 1 = ³ h 5/ 0 h 5/ µ # 5/6 1 5 µ 5/6 1 =6.15% 5 6

3. A hemisphere of radius R. Solution: V (h) = 1 3 πh (3R h) Making the right hand side equal to 3 πr3 /5 and solving for h gives the height of the 0% (remaining) volume to equal 0.391600R. Finish as your assignment. 7