1.1. Linear Constant Coefficient Equations. Remark: A differential equation is an equation


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1 Liner Constnt Coefficient Equtions Section Objective(s): Overview of Differentil Equtions. Liner Differentil Equtions. Solving Liner Differentil Equtions. The Initil Vlue Problem Overview of Differentil Equtions. Remrk: A differentil eqution is n eqution, the unknown is function, nd both the function nd its derivtives my pper in the eqution. Exmple 1.1.1: () Newton s Lw: Mss times ccelertion equls force, m = f, where m is the prticle mss, = d x/dt is the prticle ccelertion, nd f is the force cting on the prticle. Hence Newton s lw is the differentil eqution m d x dt (t) = f (t, x(t), x (t)). where the unknown is the position of the prticle in spce, x(t), t the time t. Remrk: This is second order Ordinry Differentil Eqution (ODE). (b) Rdioctive Decy: The mount u of rdioctive mteril chnges in time s follows, du (t) = k u(t), k > 0, dt where k is positive constnt representing rdioctive properties of the mteril. Remrk: This is first order ODE. (c) The Het Eqution: The temperture T in solid mteril chnges in time nd in one spce dimension ccording to the eqution T t (t, x) = k T (t, x), k > 0, x where k is positive constnt representing therml properties of the mteril. Remrk: This is first order in time nd second order in spce PDE. (d) The Wve Eqution: A wve perturbtion u propgting in time t nd in one spce dimension x through the medi with wve speed v > 0 is u t (t, x) = v u (t, x). x Remrk: This is second order in time nd spce Prtil Differentil Eqution (PDE).
2 1.1.. Liner Differentil Equtions. Definition A first order ODE on the unknown y is y (t) = f(t, y(t)), (1.1.1) where f is given nd y = dy. The eqution is liner iff the source dt function f is liner on its second rgument, y (t) = (t) y(t) + b(t). (1.1.) The liner eqution hs constnt re constnts. Otherwise the eqution hs vrible coefficients iff both nd b bove coefficients. Exmple 1.1.: () y = y + 3 is liner, constnt coefficients. (b) y = y + 4t is liner, vrible coefficients. t (c) y = 1 t y + 4t is nonliner. Exmple 1.1.3: Show tht y(t) = e t 3 is solution of the eqution y = y + 3. Solution: We need to compute the left nd righthnd sides of the eqution nd verify they gree. On the one hnd we compute y (t) = e t. On the other hnd we compute y(t) + 3 = e t 3 " + 3 = e t. We conclude tht y (t) = y(t) + 3 for ll t R.
3 Solving Liner Differentil Equtions. Theorem (Constnt Coefficients) The liner differentil eqution y = y + b, (1.1.3) with = 0, b constnts, hs infinitely mny solutions, y(t) = c e t b, c R. (1.1.4) Remrk: Eqution (1.1.4) is clled the generl solution of the differentil eqution in (1.1.3). Proof of Theorem 1.1.: First consider the cse b = 0, so y = y, with R. Then, y = y y y = ln( y ) = ln( y ) = t + c 0, where c 0 R is n rbitrry integrtion constnt, nd we used the Fundmentl Theorem of Clculus on the lst step, ln( y ) dt = ln( y ). Compute the exponentil on both sides, y(t) = ±e t+c 0 = ±e c 0 e t, denote c = ±e c 0 y(t) = c e t, c R. This is the solution of the differentil eqution in the cse tht b = 0. The cse b = 0 cn be converted into the cse bove. Indeed, " y = y + b y = y + b # " y + b # " = y + b #, since (b/) = 0. Denoting ỹ = y + (b/), the eqution bove is ỹ = ỹ. We know ll the solutions to tht eqution, ỹ(t) = c e t, c R y(t) + b = c et y(t) = c e t b. This estblishes the Theorem.
4 4 Exmple 1.1.5: Find ll solutions to the constnt coefficient eqution y = y + 3. Solution: We strt pulling common fctor on the righthnd side of the eqution, y = y + 3 " y + 3 " = y + 3 ". Denoting ỹ = y + (3/) we get ỹ = ỹ ỹ ỹ = ln( ỹ ) = ln( ỹ ) = t + c 0. We now compute exponentils on both sides, to get ỹ(t) = ±e t+c 0 = ±e t e c 0, denote c = ±e c 0, then ỹ(t) = c e t, c R. Since ỹ = y + 3, we get y(t) = c et 3, where c R.
5 The Initil Vlue Problem. Definition The initil vlue problem (IVP) is to find ll solutions y to y = y + b, (1.1.5) tht stisfy the initil condition y(0) = y 0, (1.1.6) where, b, nd y 0 re given constnts. Remrk: The differentil eqution y ssocited IVP hs only one solution. = y + b hs infinitely mny solutions, but the Theorem (Constnt Coefficients IVP) The initil vlue problem y = y + b, y(0) = y 0, for given constnts, b, y 0 R, nd = 0, hs the unique solution y(t) = y 0 + b " e t b. (1.1.7) Proof of Theorem 1.1.4: The generl solution of the differentil eqution in (1.1.5) is given in Eq. (1.1.4) for ny choice of the integrtion constnt c, y(t) = c e t b. The initil condition determines the vlue of the constnt c, s follows y 0 = y(0) = c b c = y 0 + b ". Introduce this expression for the constnt c into the differentil eqution in Eq. (1.1.5), y(t) = y 0 + b " e t b. This estblishes the Theorem.
6 6 Exmple 1.1.8: Find the unique solution of the initil vlue problem y = y + 3, y(0) = 1. (1.1.8) Solution: All solutions of the differentil eqution re given by y(t) = ce t 3, where c is n rbitrry constnt. The initil condition in Eq. (1.1.8) determines c, 1 = y(0) = c 3 c = 5. Then, the unique solution to the initil vlue problem bove is y(t) = 5 et 3.
1.2. Linear Variable Coefficient Equations. y + b "! = a y + b " Remark: The case b = 0 and a nonconstant can be solved with the same idea as above.
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