First order differential equations

First order differential equations Samy Tindel Purdue University Differential equations and linear algebra - MA 262 Taken from Differential equations and linear algebra by Goode and Annin Samy T. First order Differential equations 1 / 69

Outline 1 How differential equations arise 2 Basic ideas and terminology 3 Geometry of first order differential equations 4 Separable equations 5 Some simple population models 6 First order linear equations 7 Modeling problems using first order linear differential equations 8 Change of variables 9 Exact differential equations 10 Some higher order differential equations 11 Chapter review Samy T. First order Differential equations 2 / 69

Newton s second law of motion Quantities: For an object of mass m Force F Velocity v(t) at time t Displacement y = y(t) Newton s law: m dv dt = F Differential equation: Since v = dy, we get dt m d 2 y dt 2 = F Samy T. First order Differential equations 4 / 69

Spring force Physical system: spring-mass with no friction Hooke s law: The spring force is given by F s = k y Samy T. First order Differential equations 5 / 69

Second order differential equation Differential equation: Equation involving the derivatives of a function In particular the unknown is a function Equation for spring-mass system: According to Newton s and Hooke s laws m d 2 y = ky (1) dt2 Samy T. First order Differential equations 6 / 69

Second order differential equation (2) Equation for spring-mass system (2): Set Then (1) is equivalent to Solution: Of the form ω = k m d 2 y dt 2 + ω2 y = 0. y(t) = A cos(ωt φ). Samy T. First order Differential equations 7 / 69

Order of a differential equation Definition: Order of a differential equation = Order of highest derivative appearing in equation Examples: Second law of motion, spring: second order First order: y = 4 y 2 General form of n-th order differential equation: G(y, y,..., y (n) ) = 0 (2) Samy T. First order Differential equations 9 / 69

More vocabulary Linear equations: of the form a 0 (x)y (n) + a 1 (x)y (n 1) + + a n (x)y = F (x) Initial value problem: A differential equation G(y, y,..., y (n) ) = 0, plus initial values in order to get a unique solution: y(x 0 ) = y 0, y (x 0 ) = y 1,..., y (n 1) (x 0 ) = y n 1 General solution: When no initial condition is specified Solution given in terms of constants c 1,..., c n Samy T. First order Differential equations 10 / 69

Existence and uniqueness result Theorem 1. General nonlinear equation: Hypothesis: y = f (t, y), y(t 0 ) = y 0 R. (3) (t 0, y 0 ) R, where R = (α, β) (γ, δ). f and f y continuous on R. Conclusion: One can find h > 0 such that there exists a unique function y satisfying equation (3) on (t 0 h, t 0 + h). Samy T. First order Differential equations 12 / 69

Example of existence and uniqueness Equation considered: y = 3x y 1/3, and y(0) = a. Application of Theorem 1: we have f (x, y) = 3x y 1/3, f 2/3 (x, y) = xy y Therefore if a 0: 1 There exists rectangle R such that (0, a) R f and f y continuous on R 2 According to Theorem 1 there is unique solution on interval ( h, h), with h > 0 Samy T. First order Differential equations 13 / 69

Gravity example Gravity equation with friction dv dt = 9.8 v 5 (4) Samy T. First order Differential equations 14 / 69

Slope field Meaning of the graph: Values of dv according to values of v dt What can be seen on the graph: Critical value: v c = 49ms 1, solution to 9.8 v = 0 5 If v < v c : positive slope If v > v c : negative slope Samy T. First order Differential equations 15 / 69

General form of separable equation Definition 2. A first order separable differential equation is of the form p(y) dy dx = q(x) (5) Solving separable equations: Integrate on both sides of (5). Samy T. First order Differential equations 17 / 69

Example of separable equation Equation: (1 + y 2 ) dy dx = x cos(x) General solution: After integration by parts y 3 + 3y = 3(x sin(x) + cos(x)) + c, where c R Remark: Solution given in implicit form. Samy T. First order Differential equations 18 / 69

Example 2 of separable equation Equation: x dx + y exp( x) dy = 0, y(0) = 1 Unique solution: y(x) = (2 exp(x) 2x exp( x) 1) 1/2 Remark: Radical vanishes for x 1 1.7 and x 2 0.77 Samy T. First order Differential equations 19 / 69

Malthusian growth Hypothesis: Rate of change proportional to value of population Equation: for k R and P 0 0, dp dt = k P, P(0) = P 0 Solution: P = P 0 exp(kt) Samy T. First order Differential equations 21 / 69

Exponential growth (2) Integral curves: Limitation of model: Cannot be valid for large time t. Samy T. First order Differential equations 22 / 69

Logistic population model Basic idea: Model: where Growth rate decreases when population increases. r reproduction rate C carrying capacity ( dp dt = r 1 P ) P, (6) C Samy T. First order Differential equations 23 / 69

Logistic model: qualitative study Information from slope field: Equilibrium at P = C If P < C then t P increasing If P > C then t P decreasing Possibility of convexity analysis Samy T. First order Differential equations 24 / 69

Logistic model: solution First observation: Equation (6) is separable Integration: Integrating on both sides of (6) we get ( ) P ln = rt + c C P 1 which can be solved as: P(t) = c 2C c 2 + e rt Initial value problem: If P 0 is given we obtain P(t) = C P 0 P 0 + (C P 0 )e rt Samy T. First order Differential equations 25 / 69

Information obtained from the resolution Asymptotic behavior: lim P(t) = C t Prediction: If Logistic model is accurate P 0, r and C are known Then we know the value of P at any time t Samy T. First order Differential equations 26 / 69

Standard form Definition 3. A first order linear differential equation can be written as a(x) dy dx + b(x)y = r(x) Standard form: A linear equation should always be rewritten as dy dx + p(x)y = q(x) Samy T. First order Differential equations 28 / 69

Integrating factor method Recall: Our equation is dy dx Recipe to solve the equation: 1 Solve the separable equation + p(x)y = q(x) (7) di dx = p(x)i 2 Multiply equation (7) by I. It becomes d(i y) dx = q(x)i(x) (8) 3 Integrate (8) on both sides Samy T. First order Differential equations 29 / 69

Application Equation: dy dx + x y = xe x 2 2, y(0) = 1 Application of the method: 1 Integrating factor: 2 Solution: I(x) = e x2 2 y(x) = e x 2 2 + e x2 2 2 = cosh ( ) x 2 2 Samy T. First order Differential equations 30 / 69

Second application Equation: Standard form: x dy dx + 2y = cos(x), x > 0 dy dx + 2y x = cos(x) x Application of the method: 1 Integrating factor: I(x) = x 2 2 Solution: y(x) = x sin(x) + cos(x) + c x 2 Samy T. First order Differential equations 31 / 69

Mixing problem Physical situation: Tank initially containing V 0 liters with A 0 grams of chemical Flow in: solution with c 1 grams/liter of chemical Rate for the inflow: r 1 liters/minute Solution stirred, so that mixture is kept uniform Flow out: Mixture at rate r 2 liters/minute Quantity of interest: A(t) quantity of chemical in the tank at time t Samy T. First order Differential equations 33 / 69

Mixing problem (2) Volume: We have V (t) = V 0 + (r 1 r 2 )t Hypothesis: Variations of A due to flows in and out, da dt = rate of chemical in rate of chemical out Equation: Equation, standard form: da dt = c 1 r 1 A V r 2, A(0) = A 0 da dt + r 2 V 0 + (r 1 r 2 )t A = c 1 r 1, A(0) = A 0 Samy T. First order Differential equations 34 / 69

Example of mixing problem Physical situation: Tank with 8L of water and 32g of chemical Inflow: solution with 2g/L of chemical, rate 4L/min Outflow: rate 2L/min Questions: Amount of chemical after 20 min Concentration of chemical at that time Samy T. First order Differential equations 35 / 69

Example of mixing problem (2) Equation for the volume: Equation for A: V (t) = 2(t + 4) da dt + 1 t + 4 A = 8, A(0) = A 0 Integrating factor: I = t + 4 Samy T. First order Differential equations 36 / 69

Example of mixing problem (3) General solution: A(t) = 1 [ 4(t + 4) 2 + c ] t + 4 Initial value problem: imposing A(0) = 32g we get A(t) = 4 [ (t + 4) 2 + 16 ] t + 4 Value and concentration at time 20: and A(20) = 296 3 = 98.67g, c(20) = A(20) V (20) = 296 48 3 = 37 18 = 2.05g/L Samy T. First order Differential equations 37 / 69

General form of homogeneous equation General form of first order equation: dy dx General form of homogeneous equation: = f (x, y) (9) ( ) dy y dx = F. x How to see if an equation is homogeneous: When in (9) we have f (tx, ty) = f (x, y) Heuristics to solve homogeneous equations: Go back to a separable equation Samy T. First order Differential equations 39 / 69

Solving homogeneous equations Equation: General method: ( ) dy y dx = F. (10) x 1 Set y(x) = x V (x), and express y in terms of x, V, V. 2 Replace in equation (10) separable equation in V. 3 Solve the separable equation in V. 4 Go back to y recalling y = x V. Theorem 4. For equation (10), the function V satisfies which is a separable equation 1 F (V ) V dv = 1 x dx, Samy T. First order Differential equations 40 / 69

Example of homogeneous equation Equation: dy dx = 4x + y x 4y Equation for V : 1 4V 4(1 + V 2 ) dv = 1 x dx Solution for V : 1 4 arctan(v ) 1 2 ln(1 + V 2 ) = ln( x ) + c 1 Solution for y: ( ) 1 y 2 arctan ln ( x 2 + y 2) = c 2 (11) x Samy T. First order Differential equations 41 / 69

Example of homogeneous equation (2) Polar coordinates: Set x = r cos(θ), and y = r sin(θ) that is ) r = ( x 2 + y 2) 1/2, and θ = arctan ( y x Solution in polar coordinates: Equation (11) becomes r = c 3 e θ 4 Samy T. First order Differential equations 42 / 69

Bernoulli equations Definition 5. A Bernoulli equation is of the form y + p(x)y = q(x)y n (12) Recipe to solve a Bernoulli equation: 1 Divide equation (12) by y n 2 Change of variable: u = y 1 n 3 The equation for u is a linear equation of the form 1 1 n u + p(x)u = q(x) Samy T. First order Differential equations 43 / 69

Example of Bernoulli equation Equation: y + 3 2/3 12y y = x (1 + x 2 ) 1/2 Solution: 1 Divide the equation by y 2/3. We get y 2/3 y + 3 x y 1/3 = 12 (1 + x 2 ) 1/2 2 Change of variable u = y 1/3. We end up with the linear equation u + 1 x u = 4 (1 + x 2 ) 1/2 Samy T. First order Differential equations 44 / 69

Example of Bernoulli equation (2) Solving the linear equation: Integrating factor given by I(x) = x Then integrating we get u(x) = x ( 1 4(1 + x 2 ) 1/2 + c ) Going back to y: We have u = y 1/3. Thus y(x) = x 3 ( 4(1 + x 2 ) 1/2 + c ) 3 Samy T. First order Differential equations 45 / 69

Example of exact equation Equation considered: 2x + y 2 + 2xyy = 0 (13) Remark: equation (13) neither linear nor separable Additional function: Set φ(x, y) = x 2 + xy 2. Then: φ x = 2x + y 2, and φ y = xy. Samy T. First order Differential equations 47 / 69

Example of exact equation (2) Expression of (13) in terms of φ: we have (13) φ x + φ dy y dx = 0 Solving the equation: We assume y = y(x). Then (13) dφ (x, y) = 0 φ(x, y) = c, dx for a constant c R. Conclusion: equation solved under implicit form x 2 + xy 2 = c. Samy T. First order Differential equations 48 / 69

General exact equation Proposition 6. Equation considered: M(x, y) dx + N(x, y) dy = 0. (14) Hypothesis: there exists φ : R 2 R such that φ x = M(x, y) and φ y = N(x, y). Conclusion: general solution to (14) is given by: φ(x, y) = c, with c R, provided this relation defines y = y(x) implicitely. Samy T. First order Differential equations 49 / 69

Criterion for exact equations Notation: For f : R 2 R, set f x = f x and f y = f y Theorem 7. Let: R = {(x, y); α < x < β, and γ < y < δ}. M, N, M y, N x continuous on R. Then there exists φ such that: φ x = M, and φ y = N on R, if and only if M and N satisfy: M y = N x on R Samy T. First order Differential equations 50 / 69

Computation of function φ Aim: If M y = N x, find φ such that φ x = M and φ y = N. Recipe in order to get φ: 1 Write φ as antiderivative of M with respect to x: φ(x, y) = a(x, y) + h(y), where a(x, y) = M(x, y) dx 2 Get an equation for h by differentiating with respect to y: h (y) = N(x, y) a y (x, y) 3 Finally we get: φ(x, y) = a(x, y) + h(y). Samy T. First order Differential equations 51 / 69

Computation of φ: example Equation considered: y cos(x) + 2xe y + ( sin(x) + x 2 e y 1 ) y = 0. (15) }{{}}{{} M N Step 1: verify that M y = N x on R 2. Step 2: compute φ according to recipe. We find Solution to equation (15): φ(x, y) = y sin(x) + x 2 e y y y sin(x) + x 2 e y y = c. Samy T. First order Differential equations 52 / 69

Computation of φ: counter-example Equation considered: Step 1: verify that M y N x. 3xy + y 2 + ( x 2 + xy ) y = 0. (16) }{{}}{{} M N Step 2: compute φ according to recipe. We find h (y) = x 2 2 xy still depends on x! Conclusion: Condition M y = N x necessary. Samy T. First order Differential equations 53 / 69

Solving an exact equation: example Equation considered: 2x y + (2y x) y = 0, y(1) = 3. (17) }{{}}{{} M N Step 1: verify that M y = N x on R 2. Step 2: compute φ according to recipe. We find φ(x, y) = x 2 xy + y 2. Solution to equation (17): recalling y(1) = 3, we get x 2 xy + y 2 = 7. Samy T. First order Differential equations 54 / 69

Solving an exact equation: example (2) Expressing y in terms of x: we get y = x ( 2 ± 7 3x 2 ) 1/2. 4 Recalling y(1) = 3, we end up with: y = x ( 2 + 7 3x 2 ) 1/2. 4 Interval of definition: 7 7 x 2 4 ; 2 ( 3.05 ; 3.05) 4 Samy T. First order Differential equations 55 / 69

Integrating factor Definition 8. Let I be a non zero function. Then I is an integrating factor for (14) if the equation I(x, y)m(x, y) dx + I(x, y)n(x, y) dy = 0 is exact Samy T. First order Differential equations 56 / 69

Criterions to find integrating factors Criterion 1: We have I = I(x) iff M y N x N In this case I is given by ( I(x) = exp Criterion 2: We have I = I(y) iff In this case I is given by M y N x M I(y) = exp ( = f (x). ) f (x) dx. = g(y). ) g(y) dy. Samy T. First order Differential equations 57 / 69

Example of integrating factor Equation: ( 2x y 2 ) dx + xy dy = 0, x > 0 (18) Remark: Equation (18) is not exact, since M y N x Criterion 1: We have M y N x = 3 N x Therefore an integrating factor is ( ) 3 I(x) = exp x dx = 1 x 3 Samy T. First order Differential equations 58 / 69

Example of integrating factor (2) Exact equation: Multiplying (18) by I we get ( 2x 2 x 3 y 2) dx + x 2 y dy = 0 Solution: or equivalently 1 2 x 2 y 2 2x 1 = c y 2 4x = c 1 x 2 Samy T. First order Differential equations 59 / 69

Objective General 2nd order differential equation: d 2 ( y dx = F x, y, dy ) 2 dx Aim: See cases of 2nd order differential equations which can be solved with 1st order techniques Samy T. First order Differential equations 61 / 69

2nd order eq. with missing dependent variable Case 1: Equation of the form d 2 ( y dx = F x, dy ) 2 dx Method for case 1: the function v = y solves dv dx Then compute y = v(x) dx. = F (x, v). Samy T. First order Differential equations 62 / 69

Example Equation: d 2 y dx = 1 ( ) dy 2 x dx + x 2 cos(x), x > 0. Change of variable: v = y solves the linear equation Integrating factor: v x 1 v = x cos(x) ( I(x) = exp ) x 1 dx = x 1 Solving for v: Solving for y: y = v = x sin(x) + cx v = x cos(x) + sin(x) + c 1 x 2 + c 2 Samy T. First order Differential equations 63 / 69

2nd order eq. with missing independent variable Case 2: Equation of the form d 2 ( y dx = F y, dy ) 2 dx Method for case 2: We set v = dy. Then observe that dx d 2 y dx = dv 2 dx = dv dy Thus v solves the 1st order equation v dv dy = F (y, v). dy dx = v dv dy Samy T. First order Differential equations 64 / 69

Example Equation: d 2 y dx = 2 2 1 y ( ) 2 dy. dx Change of variable: v = y solves the 1st order separable equation v dv dy = 2 1 y v 2 Solving for v: v(y) = c 1 (1 y) 2 Samy T. First order Differential equations 65 / 69

Example (2) Separable equation for y: dy dx = c 1(1 y) 2 Solving for y: y = c 1x + (c 2 1) c 1 x + c 2 Samy T. First order Differential equations 66 / 69

Review table Type Standard form Technique Separable p(y)y = q(x) Separate variables and integrate Linear y + p(x)y = q(x) Integrating factor I = e p(x)dx Homog. y = f (x, y) where Set y = xv f (tx, ty) = f (x, y) V solves separable equation Bernoulli y + p(x)y = q(x)y n Divide by y n, set u = y 1 n u solves linear equation Exact M dx + N dy =0 Solution φ(x, y) = c, where φ with M y = N x integral of M and N Samy T. First order Differential equations 68 / 69

Example Equation: dy 5 dx = 8x + 3y 4 4xy 3 Type of method: Not separable, not homogeneous, not linear Bernoulli, under the form Not exact under the form ( 8x 5 + 3y 4) y + 3 4x y = 2x 4 y 3 } {{ } M dx + 4xy 3 dy }{{} N My Nx We have = 2x 1 N We find an integrating factor I(x) = x 2 Samy T. First order Differential equations 69 / 69