4/5/2021 Differential Equations Cont’d Second Order Differential Equation A second order differential equation can be formed when a function is differentiated twice Therefore they can be solved by integrating twice. The general solution will contain two arbitrary constants. 1 4/5/2021 Find the general solution of the differential equation π π π π ππ = ππ − ππ + ππ π2 π¦ = π₯ 3 − 6π₯ + 12 ππ₯ 2 ΰΆ± π2π¦ ππ₯ = ΰΆ± π₯ 3 − 6π₯ + 12 ππ₯ ππ₯ 2 ππ¦ π₯ 4 = − 3π₯ 2 + 12π₯ + π΄ ππ₯ 4 ΰΆ± ππ¦ π₯4 ππ₯ = ΰΆ± − 3π₯ 2 + 12π₯ + π΄ ππ₯ ππ₯ 4 ΰΆ± ππ¦ π₯4 ππ₯ = ΰΆ± − 3π₯ 2 + 12π₯ + π΄ ππ₯ ππ₯ 4 π¦= π₯5 − π₯ 3 + 6π₯ 2 + π΄π₯ + π΅ 20 2 4/5/2021 A particle moves in a straight line through a fixed point O. Its acceleration is given by π2 π¦ ππ₯ 2 = 3π‘ − 4 where t is the time in seconds after passing O, and x is the ππ₯ displacement from O. The particle reaches a point A when π‘ = 2 πππ ππ‘ = 3 ππ₯ a. Find ππ‘ π€βππ π‘ = 1 b. Find x as a function of t. π2 π₯ = 3π‘ − 4 ππ‘ 2 ΰΆ± π2π₯ ππ‘ = ΰΆ± 3π‘ − 4 ππ‘ ππ‘ 2 ππ₯ 3π‘ 2 = − 4π₯ + π΄ ππ‘ 2 ππ₯ 3π‘ 2 = − 4π₯ + 5 ππ‘ 2 a πβππ π‘ = 1 ππ₯ 3(1)2 = − 4(1) + 5 ππ‘ 2 πβππ π‘ = 2 πππ ππ₯ =3 ππ‘ 3 = −2 + π΄ π΄=5 ππ₯ 3π‘ 2 = − 4π₯ + 5 ππ‘ 2 ππ₯ 3π‘ 2 = − 4π₯ + 5 ππ‘ 2 b. ΰΆ± ππ₯ 3π‘ 2 ππ‘ = ΰΆ± − 4π₯ + 5ππ‘ ππ‘ 2 π₯= π‘3 − 2π₯ 2 + 5π₯ + π΅ 2 ππ₯ 5 = ππ −1 ππ‘ 2 3 4/5/2021 Suppose we have the first order differential equation ππ¦ + ππ¦ = π ππ₯ Example: Solve ππ¦ ππ₯ + 3π¦ π₯ ππ₯ = π₯3 This cannot be solved by ‘regular’ differential equations Recall π¦ = π’π£ ΰΆ± π£ππ’ + π’ππ£ ππ₯ = π’π£ ππ¦ π’π£ = π£ππ’ + π’ππ£ ππ₯ 4 4/5/2021 Integrating Factor ππ¦ + ππ¦ = π ππ₯ We multiply both sides of the differential equation by the integrating factor I which is defined as πΌ = π β«π₯ππ Χ¬β¬ Multiplying our original differential equation by I we get that: ππ¦ + ππ¦ = π ππ₯ →πΌ ππ¦ + πΌππ¦ = πΌπ ππ₯ → ΰΆ±πΌ ππ¦ + πΌππ¦ ππ₯ = ΰΆ± πΌπ ππ₯ ππ₯ Product Rule → πΌπ¦ = ΰΆ± πΌπ ππ₯ 5 4/5/2021 Find the general solution of the differential equation ππ¦ + 2π¦ = π −π₯ ππ₯ π 2π₯ πΌ = π β«π₯ππ Χ¬β¬ πΌ = π β« Χ¬β¬2ππ₯ ΰΆ± π 2π₯ ππ¦ + 2π 2π₯ π¦ = π 2π₯ β π −π₯ ππ₯ ππ¦ + 2π 2π₯ π¦ ππ₯ = ΰΆ± π π₯ ππ₯ ππ₯ π¦π 2π₯ = π π₯ + π πΌ = π 2π₯ π¦ = π π₯ β π −2π₯ + π −2π₯ π π¦ = π −π₯ + π −2π₯ π Find the general solution of the differential equation ππ¦ 3 ππ₯ + π¦= 3 ππ₯ π₯ π₯ ππ¦ 3 ππ₯ + π¦= 3 ππ₯ π₯ π₯ π₯3 ππ¦ + 3π₯ 2 π¦ = π π₯ ππ₯ 3 πΌ = π β«π₯ππ₯Χ¬β¬ πΌ = π 3πππ₯ πΌ = π πππ₯ πΌ = π₯3 ΰΆ± π₯3 ππ¦ + 3π₯ 2 π¦ ππ₯ = ΰΆ± π π₯ ππ₯ π₯3π¦ = π π₯ + π 3 π¦= ππ₯ + π π₯3 6 4/5/2021 Find the general solution of the differential equation ππ¦ 2 − 4π¦ = 1 ππ₯ ππ¦ 1 − 2π¦ = ππ₯ 2 πΌ = π β« Χ¬β¬−2ππ₯ π −2π₯ ππ¦ 1 − 2π −2π₯ π¦ = π −2π₯ ππ₯ 2 ΰΆ± π −2π₯ πΌ = π −2π₯ ππ¦ 1 − 2π −2π₯ π¦ ππ₯ = ΰΆ± π −2π₯ ππ₯ ππ₯ 2 π −2π₯ π¦ = − π¦= π −2π₯ +π 4 1 + π 2π₯ π 4 “ Homogeneous Linear Differential Equations 7 4/5/2021 π2 π¦ ππ¦ To solve differential equations ππ¦ ′′ + ππ¦ ′ + ππ¦ = 0 (π ππ₯2 + π ππ₯ + ππ¦ = 0) Find the roots πΌ πππ π½ of the auxiliary equation ππ’2 + ππ’ + π = 0 If πΌ πππ π½ are real and distinct π 2 > 4ππ , the general solution is π¦ = π΄π πΌπ₯ + π΅π π½π₯ If πΌ πππ π½ are real but equal π 2 = 4ππ , the general solution is π¦ = π πΌπ₯ (π΄ + π΅π₯) If πΌ πππ π½ are complex π 2 < 4ππ , the general solution is π¦ = π ππ₯ (π΄ cos ππ₯ + π΅ sin ππ₯) Find the general solution of the differential equation π¦ ′′ − 4′ + 4π¦ = 0 π’2 − 4π’ + 4 = 0 π’−2 2 =0 If πΌ πππ π½ are real but equal π 2 = 4ππ , the general solution is π¦ = π πΌπ₯ (π΄ + π΅π₯) π¦ = π 2π₯ (π΄ + π΅π₯) 8 4/5/2021 Find the general solution of the differential equation π¦ ′′ + 3′ − 10π¦ = 0 π’2 + 3π’ − 10 = 0 π’+5 π’−2 =0 πΌ = −5 πππ π½ = 2 If πΌ πππ π½ are real and distinct π 2 > 4ππ , the general solution is π¦ = π΄π πΌπ₯ + π΅π π½π₯ π¦ = π΄π −5π₯ + π΅π 2π₯ “ Nonhomogeneous Linear Differential Equations 9 4/5/2021 ππ¦ ′′ + ππ¦ ′ + ππ¦ ≠ 0 is a nonhomogeneous second order differential equation If the right hand side of the equation is not 0, then the particular solution can be found as follows: First, find the form of the solution of the corresponding homogeneous equation keeping the constants A and B as such: this is called the complementary solution π¦π (π₯) Second, find a particular integral of the ODE yπ (π₯) Then the solutions of the ODE are of the form: π¦(π₯) = π¦π (π₯) + π¦π (π₯) Undetermined Coefficients This method consists in making an educated guess as to what the particular integral should look like. The following table can be used: F(x) π ππ πππ πππππ ππ πππππ Particular Integral πͺ πͺπ + π« πͺππ + π«π + π¬ πͺππππ + π«ππππ 10 4/5/2021 Find the general solution of the differential equation π¦ ′′ − 2π¦ ′ + 5π¦ = 10π₯ + 1 π’2 − 2π’ + 5π¦ = 0 π’= 2 ± 4 − 20 2 π’ = 1 ± 2π If πΌ πππ π½ are complex π 2 < 4ππ , the general solution is π¦ = π ππ₯ (π΄ cos ππ₯ + π΅ sin ππ₯) π¦ = π π₯ (π΄πππ 2π₯ + π΅ sin 2π₯) Finding the particular Integral π¦ ′′ − 2π¦ ′ + 5π¦ = 10π₯ + 1 π¦ = πΆπ₯ + π· π¦′ = πΆ π¦′′ = 0 ∴ 0 − 2πΆ + 5(πΆπ₯ + π·) = 10π₯ + 1 ∴ 5πΆπ₯ + 5π· − 2πΆ = 10π₯ + 1 5πΆπ₯ = 10π₯ πΆ=2 5π· − 2πΆ = 1 π·=1 ππΌ = 2π₯ + 1 11 4/5/2021 π¦(π₯) = π¦π (π₯) + π¦π (π₯) π¦(π₯) = π π₯ (π΄πππ 2π₯ + π΅ sin 2π₯) + 2π₯ + 1 Find the general solution of the differential equation π¦ ′′ − 2π¦ ′ = 5 sin π₯ π’2 − 2π’ = 0 π’ π’−2 =0 πΌ = 0 πππ π½ = 2 If πΌ πππ π½ are real and distinct π 2 > 4ππ , the general solution is π¦ = π΄π πΌπ₯ + π΅π π½π₯ π¦ = π΄ + π΅π 2π₯ 12 4/5/2021 Finding the particular Integral π¦ ′′ − 2π¦ ′ = 5 sin π₯ π¦ = πΆπππ π₯ + π·π ππ π₯ π¦ ′ = −πΆπ πππ₯ + π·πππ π₯ π¦ ′′ = −πΆπππ π₯ − π·π πππ₯ ∴ −πΆπππ π₯ − π·π πππ₯ − 2 −πΆπ πππ₯ + π·πππ π₯ = 5 sin π₯ ∴ πππ π₯ −πΆ − 2π· + π πππ₯ 2πΆ = π· = 5π πππ₯ −πΆ − 2π· = 0 πΆ = −2π· 2πΆ − π· = 5 π· = −1 ππΌ = 2 cos π₯ − sin π₯ πΆ=2 π¦(π₯) = π¦π (π₯) + π¦π (π₯) π¦ π₯ = π΄ + π΅π 2π₯ + 2 cos π₯ − sin π₯ 13
