Partial preview of the text
Download First Order Ordinary Differential Equations: Concepts, Methods, and Solved Examples and more Study Guides, Projects, Research Differential and Integral Calculus in PDF only on Docsity!
Chapter 2 First Order Ordinary Differential Equations The complexity of solving Dk’s increases with the order. We begin with first order DE’s. 2.1 Separable Equations A first order ODE has the form F(a, y,y’) =0. In theory, at least, the methods of algebra can be used to write it in the form™ y! = G(w,y). If G(w,y) can be factored to give G(a,y) = M(x) N(y),then the equation is called separable. M(x) N(y), we rewrite it in the form To solve the separable equation y! (yy! = g(x). Integrating both sides gives [tors de= foo), fina [ro de. Example 2.1. Solve 2xy +6a+ (a? —4) y' =D. * “We use the notation dy/de = G(x, y) and dy = G(2, y) dx interchangeably. 5 6CHAPTER 2. FIRST ORDER ORDINARY DIFFERENTIAL EQUATIONS Solution. Rearranging, we have (x? — 4) y' = —2ay — 62, = —2cy — 6a, i, te yt3 a4! In(\y + 3|) = —In(|a? —4]) + C. In(|y +3) +In(|2? - 4|) =C, a#t2 where C is an arbitrary constant. Then |v+3) (@?-4)| = 4 (y +3) (w? — 4) = A, 3 ' yt =a where A is a constant (equal to +e°) and a 4 +2. Also y = —3 is a solution (corresponding to A = 0) and the domain for that solution is R. Oy Example 2.2. Solve the vp sin(«) de + y dy =0, where y(0) =1. « Solution. Note: sin(w) dw + ydy = 0 is an alternate notation meaning the same as sin(«) + ydy/dx = 0. We have y dy = —sin(x) de, [vta= [ -sin(oyae, v 5 = cose) + C1, y= /2eos(e) + Ca, where C; is an arbitrary constant and Cz =2C;. Considering y(0) =1, we have 1= 7240) =)1=240—>Q=-1. Therefore, y = \/2cos(x) — 1 on the domain (—7/3, 7/3), since we need cos(x) > 1/2 and cos(+n/3) = 1/2. 8CHAPTER 2. FIRST ORDER ORDINARY DIFFERENTIAL EQUATIONS Theorem 2.4 If F and G are functions that are continuously differentiable throughout a simply connected region, then F da + G dy is exact if and only if @G/Ox = OF /dy. Proof. Proof is given in MATB42. a Example 2.5. Consider (3x74 + 2?) da + (2a°y + y*) dy =0. Let w = (307y? + 2) de + (20°y + y?) dy ae Then note that By THEOREM 2.4, w = dg for some g. To find g, we know that ao 3a7y? + 2, (2.1a) D 3 = 2Qe8y ty? (2.1b) Integrating Equation (2.1a) with respect to a gives us — 3,2 a "7 g= 8 +> +hly). (2.2) So differentiating that with respect to y gives us Eq, (2.1b) a dh 2 93,4 ay > 2ay + dy’ ; dh as y + y? = wy +S, dy dh yy y, 3 hy) =“H+e 3 2.2. EXACT DIFFERENTIAL EQUATIONS 9 for some arbitrary constant C. Therefore, Equation (2.2) becomes Note that according to our differential equation, we have a ya 3 av eee +c) =O which implies «4? +4 4 40=C" for some arbitrary constant C’. Letting D = C’— C, which is still an arbitrary constant, the solution is 3 3,2. 2 ¥ z+yZ =D. By ch ety Example 2.6. Solve (3a? + 2xy?) da + (22*y) dy = 0, where y(2)=—3. — * Solution. We have / (32? +2xy?) de = 03 +224? +0 for some arbitrary constant C’. Since C’ is arbitrary, we equivalently have x? + a2y? = CO. With the initial condition in mind, we have 84+4-9=C=3CH=44 Therefore, x3 + x?y? = 44 and it follows that tV44—23 y= 2 @ But with the restriction that y(2) = —3, the only solution is on the domain (— 9/44, 4A) \ {0}. ° Let w = Fdv + Gdy. Let y = s(a) be the solution of the DE w = 0, ie., F+Gs'(x) =0. Let yo = s(a) and let 7 be the piece of the graph of y = s(x) from (2, yo) to («,y). Figure 2.1 shows this idea. Since y = s() is a solution tow =0, we must have w = 0 along +. Therefore, fw = 0. This can be soon 2.3. INTEGRATING FACTORS i So the equation is exact. © 2.3 Integrating Factors Consider the equation w = 0. Even if w is not exact, there may be a function I(x, y) such that Iw is exact. Sow =0 can be solved by multiplying both sides by I. The function J is called an integrating factor for the equation w = 0. Example 2.7. Solve y/x? + 1+ y'/« =0. * Solution. We have , (541) ae+ = ay =o. zs & [a (=)=-al¢la-a (279): So the equation is not exact. Multiplying by x? gives us We see that (y+?) da+ady =0, - slay) 0, eo He ryt s for some arbitrary constant C. Solving for y finally gives us © There is, in general, no algorithm for finding integrating factors. But the following may suggest where to look. It is important to be able to recognize common exact forms: ady +ydx=d(ay), ad: wdy—ydx =a(), we z wdw + ydy AS +2 "), =d ety awdy—ddv _ ety =a (tan ( *)): at ty! (ay de + be dy) = d(27y’) . 12CHAPTER 2. FIRST ORDER ORDINARY DIFFERENTIAL EQUATIONS Example 2.8. Solve (xy? + y) dx + (2x*y — x) dy =0. * Solution. Expanding, we have 2? da + 2a? y dy + ydx — x dy =0 Here, a= 1 and b =2. Thus, we wish to use d(ary?) = y? dex + 2ay dy. This suggests dividing the original equation by «? which gives dx — x dy y Pde + xy dy +2 A — Therefore, wy +4=c, «#0, ¢ where C’ is an arbitrary constant. Additionally, y= 0 on the domain R is a gOlaHod RETO tigiial €guRtiOn! ° Example 2.9. Solve ydx — x dy — (a? + y?) de =0. * Solution. We have yde —edy deco or unless « = 0 and y = 0. Now, it follows that —tan7 ‘O- e =C, ©) (=? =tan(D—<), ¥ v y=wtan(D— 2), where C is an arbitrary constant and the domain is D-2#(n+1)F, eé(Qn+1) 2 for any integer n. Also, since the derivation of the solution is based on the assumption that « # 0, it is unclear whether or not 0 should be in the domain, ie., does y = x tan(D — 2) satisfy the equation when x = 0? We have y— ay! — 14CHAPTER 2. FIRST ORDER ORDINARY DIFFERENTIAL EQUATIONS If we can find any particular solution I(«,y) of the PDE IN + INg = IyM +IMy, (#*) then we can use it as an integrating factor to find the general solution of (+). Unfortunately, (4) is usually even harder to solve than (+), but as we shall see, there are times when it is easier. Example 2.12. We could look for an I having only e's and no y's? For exam- ple, consider J, = 0. Then I, InN + INy = IMy implies = = This works if (My — Nz) /N happens to be a function of x alone. Then Pel Magh de Similarly, we can also reverse the role of w and y. If (N; — My) /M happens to be a function of y alone, then [75 t ay e! works. * 2.4 Linear First Order Equations A first order linear equation (n = 1) looks like y' + P(a)y = Q(2). An integrating factor can always be found by the following method. Consider dy + P(a)y dx = Q(x) dx, (P(x)y — Q(x)) de + dy =0. —_——a Ww M(ry) N(x.) We use the DE for the integrating factor I(v,y). The equation IM dx + IN dy is exact if I,N + IN, =1yM+IM,. 2.4. LINEAR FIRST ORDER EQUATIONS 15 In our case, I, +0= I, (P(a)y — Q(a)) + IP(2). (*) We need only one solution, so we look for one of the form I(x), ie., with Iy = 0. Then (+) becomes dl & =IP(a). This is separable. So ¢ = P(x) de, In(7|) = [Pe dx +C. =elP@e, & >0 T= el Pie) de, We conclude that ef P(#)4” js an integrating factor for y/ + P(x)y = Q(z). where x > 0. * Example 2.13. Solve y!—(1/r)y =< Solution. Here P(x) = —1/x. Then Pa el Plo) de p-fdde — g-in(lepae 1 _ where # > 0. Our differential equation is wdy ~ yde Bey USE oS de: x Multiplying by the integrating factor 1/a gives us dy — yy de _ 92 dp z Then 3 ye Jo 46, x 37 3 y= Zt Or on the domain (0, 00), where C'is an arbitrary constant (« > Ois given). 2.5, SUBSTITUTIONS 17 where J = ef ?4Y = ¢?/, Therefore, cu 4 270%) =e, dy xe =—e4+C, where C is an arbitrary constant. We could solve explicitly for y, but it is messy. The domain is not easy to determine. > 2.5 Substitutions In many cases, equations can be put into one of the standard forms discussed above (separable, linear, etc.) by a substitution. Example 2.16. Solve y’” — 2y! =5. * Solution. This is a first order linear equation for y!. Let 1 = y!. Then the equation becomes ul —2u=5. The integration factor is then J = e~/?4 = e~?*, Thus, ene 20 w — Que~?* = Be 2 5 _s e727 = se" +C, where C is an arbitrary constant. But w= y/, so 3 5 eF+C,= —3et Cye* + Co on the domain R, where C; and C) are arbitrary constants. © We now look at standard substitutions. 2.5.1 Bernoulli Equation The Bernoulli equation is given by dy 9 = on 7 Te + Poly = O(@)y"- Let z= y!-". Then dz | _ndy mote a 18CHAPTER 2. FIRST ORDER ORDINARY DIFFERENTIAL EQUATIONS giving us y+ Play" = Qe), oS + Plws = at, © £m) Ple)s = (1) Q@), which is linear in 2. Example 2.17. Solve y! + ey =«y?. * Solution. Here, we have n = 3. Let z = y~?. Ify #0, then dz _y,-1dy de de Therefore, our equation becomes ay — tay = ay, ea 2 —Z tay =2, 2! — Iny = — 2x. We can readily see that [ = e~S2*4" = e-*". Thus, * = 20", 2? =e" 46, ze=1l+ Ce’, where C is an arbitrary constant. But z = y~?. So 1 =f, oe Vite The domain is R, C>-l, lel > Von), C<-1. An additional solution is y = 0 on R. ce) 20CHAPTER 2. FIRST ORDER ORDINARY DIFFERENTIAL EQUATIONS This suggests that v = y/z (or equivalently, y = vx) might help. In fact, write _M(@,y) = r(2). Cen ae Then = R(w) Therefore, of = Riv) —» dv _ de Rs @" which is separable. We conclude that if M/ and N are homogeneous of the same degree, setting y = va will give a separable equation in v and a. Example 2.21. Solve ay? dy = (x? + y*) de. * Solution. Let y= va. Then dy = vdx + xdv, and our equation becomes ate? (vde+ adv) = («4 + v%2?) de, xu? da + xv? dv = a8 da +825 da. Therefore, x = 0 or v? dv = dx/a. So we have 3 = In(|2|) + C = In({a|) + In([A]) = In(|Az) = In(Ae). Cc where the sign of A is the opposite of the sign of a. Therefore, the general solution is y = a(3In(Az))!/%, where A is a nonzero constant. Every A > 0 yields a solution on the domain (0,00); every A < 0 yields a solution on (—00, 0). In addition, there is the solution y = 0 on the domain R. 0) 2.5.3 Substitution to Reduce Second Order Equations to First Order A second order DE has the form Fly",y',y,x) = 0. 2.5. SUBSTITUTIONS 21 If it is independent of y, namely, F(y", y/,2) = 0, then it is really just a first order equation for y! as we saw in earlier examples. Consider now the case where it is independent of x, namely, F(y", Substitute v = dy/dx for z, ie., eliminate « between the equations dy dy F(—4, 2 y) =0 (=. tu) and v = dy/dx. Then Py dv _ dvdy_ dv SG x, nf OU OO de? du dydx dy iy ay it (33h) =0~ F( Gene) This is a first order equation in v and y. Therefore, Example 2.22. Solve y= 4(y')*/7y. Solution. Let v= dy/dx. Then dy de ay, dx? dy and our equation becomes oO =4ydy, v>0, vv = dy? + Cr, Vo=y? + Cr, yy) =0. 2.5. SUBSTITUTIONS 23 will give us some cancellation. Thus, u’ = s'e” + se™ and our equation becomes ase” +ase* + se?” — ese® =e", as! + se” =e' = 7 - [Se 2 Working our way back through the subsitutions we find that s = zae~* so our 1 1+ 2a f e* ph([ ESS ) =| eo Using algebra, we could solve this equation for z in terms of « and then integrate solution becomes the result to get v which then determines y = e” as a function of 2. The algebraic solution for z is messy and we will not be able to find a closed form expression for the antidervative v, so we will abandon the calculation at this point. In practical applications one would generally use power series techniques on equations of this form and calculate as many terms of the Taylor series of the solution as are needed to give the desired accuracy. > Next consider equations of the form (are + bry + e1) da + (ax + bay + c2) dy = 0. If c, = cg = 0, the equation is homogeneous of degree 1. If not, try letting @=a—hand gj = y—k. We try to choose h and k to make the equation homogeneous. Since h and k are constants, we have dz = dx and dy = dy. Then our equation becomes (a,% +ayh + byy + bik + €1) d% + (a2% + agh + boy + bok + c2) dy = We want ah +bjk = —cy and agh + bok = —c2. We can always solve for h and k, unless a h|_o az be 24CHAPTER 2. FIRST ORDER ORDINARY DIFFERENTIAL EQUATIONS So suppose a, by =0 az be Then (a2,b2) = m(ai,bi). Let 2 = ax + bry. Then dz = ay de + by dy. If by £0, we have i naeke dy = bh (2-ter)de+ (ms Hen) 25 =, 1 a mz +2 _ (ora r®)aes (ME) ano bdr = mete 2+e+a;/b, This is a separable equation. If by =0 but by 4 0, we use z = aga + boy instead. Finally, if both b) = 0 and by = 0, then the original equation is separable 
