Calculus 3.6: solutions of first order differential equations

Infinite Series and Differential Equations Nguyen Thieu Huy Hanoi University of Science and Technology Nguyen Thieu Huy (HUST) Infinite Series and Diff Eq 1/9 Exact equations 3.1 Definition A differential equation in differential form M(x, y )dx + N(x, y )dy = (1) is exact if there exists a function g (x, y ) such that dg (x, y ) = M(x, y )dx + N(x, y )dy 3.2 Test for exactness: If M(x, y ) and N(x, y ) are continuous functions and have continuous first partial derivatives on some rectangle of the xy -plane, ) ) then Equation (1) is exact if and only if ∂M(x,y = ∂N(x,y in this domain ∂y ∂x 3.3 Solution: To solve Equation (1), first solve the equations ∂g (x,y ) = M(x, y ) ∂x for g (x, y ) ∂g (x,y ) = N(x, y ) ∂y The solution to (1) is g (x, y ) = c where c: arbitrary constant Nguyen Thieu Huy (HUST) Infinite Series and Diff Eq 2/9 3.4 Integrating Factors: In general, Equation (1) is not exact Definition A function I (x, y ) = is an integrating factor for (1) if the equation I (x, y )[M(x, y )dx + N(x, y )dy ] = (2) is exact Nguyen Thieu Huy (HUST) Infinite Series and Diff Eq 3/9 3.4 Integrating Factors: In general, Equation (1) is not exact Definition A function I (x, y ) = is an integrating factor for (1) if the equation I (x, y )[M(x, y )dx + N(x, y )dy ] = (2) is exact A solution to (1) is obtained by solving the exact differential equation (2) )M(x,y )) )N(x,y )) To find I (x, y ) we solve ∂(I (x,y∂y = ∂(I (x,y∂x Nguyen Thieu Huy (HUST) Infinite Series and Diff Eq 3/9 3.4 Integrating Factors: In general, Equation (1) is not exact Definition A function I (x, y ) = is an integrating factor for (1) if the equation I (x, y )[M(x, y )dx + N(x, y )dy ] = (2) is exact A solution to (1) is obtained by solving the exact differential equation (2) )M(x,y )) )N(x,y )) To find I (x, y ) we solve ∂(I (x,y∂y = ∂(I (x,y∂x This can be done in some special situations, results are listed in the following: Nguyen Thieu Huy (HUST) Infinite Series and Diff Eq 3/9 Example: Solve ydx − xdy = (not exact) Check ) ∂M(x,y ) − ∂N(x,y = − x2 ; Integrating factor I (x) = e − N ∂y ∂x The Eq ⇔ y dx x2 − x1 dy = which is exact The solution is Nguyen Thieu Huy (HUST) Infinite Series and Diff Eq dx x y x = x2 = C 4/9 Example: Solve ydx − xdy = (not exact) Check ) ∂M(x,y ) − ∂N(x,y = − x2 ; Integrating factor I (x) = e − N ∂y ∂x The Eq ⇔ xy2 dx − x1 dy = which is exact The solution is Linear Equations dx x y x = x2 = C 4.1 Definition A first-order linear differential equation has the form y + p(x)y = q(x) Example Looking back: Nguyen Thieu Huy (HUST) dv dt v +km = g; dI dt + Infinite Series and Diff Eq R LI = E L 4/9 Example: Solve ydx − xdy = (not exact) Check ) ∂M(x,y ) − ∂N(x,y = − x2 ; Integrating factor I (x) = e − N ∂y ∂x The Eq ⇔ xy2 dx − x1 dy = which is exact The solution is Linear Equations dx x y x = x2 = C 4.1 Definition A first-order linear differential equation has the form y + p(x)y = q(x) v dI R E Example Looking back: dv dt + k m = g ; dt + L I = L 4.2 Method of Solutions: We write in differential form ) ) = p(x) (p(x)y − q(x))dx + dy = Check N1 ∂M(x,y − ∂N(x,y ∂y ∂x Integrating factor: I (x) = e p(x)dx = When multiplied by I (x), the resulting equation e p(x)dx y + e p(x)dx p(x)y = e p(x)dx q(x) is exact Nguyen Thieu Huy (HUST) Infinite Series and Diff Eq 4/9 Example: Solve ydx − xdy = (not exact) Check ) ∂M(x,y ) − ∂N(x,y = − x2 ; Integrating factor I (x) = e − N ∂y ∂x The Eq ⇔ xy2 dx − x1 dy = which is exact The solution is Linear Equations dx x y x = x2 = C 4.1 Definition A first-order linear differential equation has the form y + p(x)y = q(x) v dI R E Example Looking back: dv dt + k m = g ; dt + L I = L 4.2 Method of Solutions: We write in differential form ) ) = p(x) (p(x)y − q(x))dx + dy = Check N1 ∂M(x,y − ∂N(x,y ∂y ∂x Integrating factor: I (x) = e p(x)dx = When multiplied by I (x), the resulting equation e p(x)dx y + e p(x)dx p(x)y = e p(x)dx q(x) is exact This equation can be solved by the method described previously However, we can proceed in a simpler way by rewriting it as d (e p(x)dx y ) = e p(x)dx q(x) dx Nguyen Thieu Huy (HUST) Infinite Series and Diff Eq 4/9 Example: Solve ydx − xdy = (not exact) Check ) ∂M(x,y ) − ∂N(x,y = − x2 ; Integrating factor I (x) = e − N ∂y ∂x The Eq ⇔ xy2 dx − x1 dy = which is exact The solution is Linear Equations dx x y x = x2 = C 4.1 Definition A first-order linear differential equation has the form y + p(x)y = q(x) v dI R E Example Looking back: dv dt + k m = g ; dt + L I = L 4.2 Method of Solutions: We write in differential form ) ) = p(x) (p(x)y − q(x))dx + dy = Check N1 ∂M(x,y − ∂N(x,y ∂y ∂x Integrating factor: I (x) = e p(x)dx = When multiplied by I (x), the resulting equation e p(x)dx y + e p(x)dx p(x)y = e p(x)dx q(x) is exact This equation can be solved by the method described previously However, we can proceed in a simpler way by rewriting it as d (e p(x)dx y ) = e p(x)dx q(x) ⇔ e p(x)dx y = e p(x)dx q(x)dx + C The dx general solution is y = e − p(x)dx e p(x)dx q(x)dx + C ∀ constant C Nguyen Thieu Huy (HUST) Infinite Series and Diff Eq 4/9 Example Solve y + x4 y = x For p(x) = y = e− 4dx x e 4dx x x dx + C = x5 + C x4 x and q(x) = x , we obtain ∀ constant C Bernoulli Equations A Bernoulli differential equation has the form y + p(x)y = q(x)y α where α is a real number (α = 0; α = 1) If α > 0, then y = is a solution Otherwise, if α < 0, then the condition is y = We now find the solutions y = To this, we divide both sides by y α to obtain y −α y + p(x)y 1−α = q(x) The substitution z = y 1−α (so, z = (1 − α)y −α y ) now transforms Eq into a linear differential equation z + (1 − α)p(x)z = (1 − α)q(x) Example: Solve y + xy = xy This is a Bernoulli differential equation with p(x) = q(x) = x, and α = First, y = is a solution To find the solution y = Dividing by y we have y −2 y + xy −1 = x Subs z = y −1 , ⇔ y = 1/z, then z = −y −2 y Subs into above equation: z −xz = −x (linear Eq) Nguyen Thieu Huy (HUST) Infinite Series and Diff Eq 5/9 Example Solve y + x4 y = x For p(x) = y = e− 4dx x e 4dx x x dx + C = x5 + C x4 x and q(x) = x , we obtain ∀ constant C Bernoulli Equations A Bernoulli differential equation has the form y + p(x)y = q(x)y α where α is a real number (α = 0; α = 1) If α > 0, then y = is a solution Otherwise, if α < 0, then the condition is y = We now find the solutions y = To this, we divide both sides by y α to obtain y −α y + p(x)y 1−α = q(x) The substitution z = y 1−α (so, z = (1 − α)y −α y ) now transforms Eq into a linear differential equation z + (1 − α)p(x)z = (1 − α)q(x) Example: Solve y + xy = xy This is a Bernoulli differential equation with p(x) = q(x) = x, and α = First, y = is a solution To find the solution y = Dividing by y we have y −2 y + xy −1 = x Subs z = y −1 , ⇔ y = 1/z, then z = −y −2 y Subs into above equation: z −xz = −x (linear Eq) ⇔ z = e− (−x)dx e (−x)dx (−x)dx x2 + C = + C e So, y = 1+C e Nguyen Thieu Huy (HUST) Infinite Series and Diff Eq x2 5/9 VI Modeling: Three main steps for modeling: Construction of the Model This involves a translation of the physical situation into mathematical terms: (A) To choose the (physical, natural, ) laws governing the problem (B) To choose mathematical variables (C) To establish the (differential) equations Nguyen Thieu Huy (HUST) Infinite Series and Diff Eq 6/9 VI Modeling: Three main steps for modeling: Construction of the Model This involves a translation of the physical situation into mathematical terms: (A) To choose the (physical, natural, ) laws governing the problem (B) To choose mathematical variables (C) To establish the (differential) equations Analysis of the Model (A) To solve equations (B) To find the properties of solutions (C) If it is too difficult to solve the equations, try to approximate the solutions (numerical methods) Nguyen Thieu Huy (HUST) Infinite Series and Diff Eq 6/9 VI Modeling: Three main steps for modeling: Construction of the Model This involves a translation of the physical situation into mathematical terms: (A) To choose the (physical, natural, ) laws governing the problem (B) To choose mathematical variables (C) To establish the (differential) equations Analysis of the Model (A) To solve equations (B) To find the properties of solutions (C) If it is too difficult to solve the equations, try to approximate the solutions (numerical methods) Comparison with Experiment or Observation (A) To interpret the information from solutions (obtained above) in the context in which the problem arose (B) To check that the mathematical solution appears physically reasonable (C) To calculate the values of the solution at selected points and compare them with experimentally observed values Or, ask whether the behavior of the solution after a long time is consistent with observations Nguyen Thieu Huy (HUST) Infinite Series and Diff Eq 6/9 Example RL-circuit: 1st Step: Modeling Physical laws: 1) Kirchhoff’s voltage law (KVL): The algebraic sum of all the instantaneous voltage drops around any closed loop is zero, or the voltage impressed on a closed loop is equal to the sum of the voltage drops in the rest of the loop.: VR + VL = E (t) 2) Ohm’s law: The voltage drop VR across a resistor is proportional to the current I : VR = RI ; 3) Voltage drop across an inductor: The voltage drop VL across an inductor is proportional to the instantaneous time rate of change of the dI ; current I: VL = L dt Nguyen Thieu Huy (HUST) Infinite Series and Diff Eq 7/9 Example RL-circuit: 1st Step: Modeling Physical laws: 1) Kirchhoff’s voltage law (KVL): The algebraic sum of all the instantaneous voltage drops around any closed loop is zero, or the voltage impressed on a closed loop is equal to the sum of the voltage drops in the rest of the loop.: VR + VL = E (t) 2) Ohm’s law: The voltage drop VR across a resistor is proportional to the current I : VR = RI ; 3) Voltage drop across an inductor: The voltage drop VL across an inductor is proportional to the instantaneous time rate of change of the dI ; current I: VL = L dt RI + L Nguyen Thieu Huy (HUST) dI dI R E (t) = E (t) ⇔ + I = dt dt L L Infinite Series and Diff Eq 7/9 dI + RL I = E L(t) Using formula dt y = e − p(x)dx e p(x)dx q(x)dx + C with x = t, y = I , p = R/L := α, and q = E /L we get 2nd Step: Solution to the equation Rt I (t) = e where α := − Rt L e L E (t) dt + C = e −αt L e αt E (t) dt + C L R L Nguyen Thieu Huy (HUST) Infinite Series and Diff Eq 8/9 dI + RL I = E L(t) Using formula dt y = e − p(x)dx e p(x)dx q(x)dx + C with x = t, y = I , p = R/L := α, and q = E /L we get 2nd Step: Solution to the equation Rt I (t) = e − Rt L e L E (t) dt + C = e −αt L e αt E (t) dt + C L where α := RL 3rd Step Case A: Constant electromotive force E = E0 αt I (t) = e −αt e RE0 + C = ER0 + C e −αt Nguyen Thieu Huy (HUST) Infinite Series and Diff Eq 8/9 dI + RL I = E L(t) Using formula dt y = e − p(x)dx e p(x)dx q(x)dx + C with x = t, y = I , p = R/L := α, and q = E /L we get 2nd Step: Solution to the equation Rt I (t) = e − Rt L e L E (t) dt + C = e −αt L e αt E (t) dt + C L where α := RL 3rd Step Case A: Constant electromotive force E = E0 αt I (t) = e −αt e RE0 + C = ER0 + C e −αt The last term tends to zero as t → ∞; practically, after some time the current I (t) will be nearly constant, equal to ER0 , the value it would have immediately (by Ohm’s law) had we no inductor in the circuit, and we see that this limit is independent of the initial value I (0) Nguyen Thieu Huy (HUST) Infinite Series and Diff Eq 8/9 dI + RL I = E L(t) Using formula dt y = e − p(x)dx e p(x)dx q(x)dx + C with x = t, y = I , p = R/L := α, and q = E /L we get 2nd Step: Solution to the equation Rt I (t) = e − Rt L e L E (t) dt + C = e −αt L e αt E (t) dt + C L where α := RL 3rd Step Case A: Constant electromotive force E = E0 αt I (t) = e −αt e RE0 + C = ER0 + C e −αt The last term tends to zero as t → ∞; practically, after some time the current I (t) will be nearly constant, equal to ER0 , the value it would have immediately (by Ohm’s law) had we no inductor in the circuit, and we see that this limit is independent of the initial value I (0) Case B: Periodic electromotive force E = E0 sin ωt For this E (t) we have that E0 α sin ωt − ω cos ωt I (t) = e −αt EL0 e αt sin ωtdt + C = C e −αt + L α2 + ω Nguyen Thieu Huy (HUST) Infinite Series and Diff Eq 8/9 Hence, I (t) = C e −αt + C e −αt + √ E0 L α sin ωt − ω cos ωt α2 + ω = E0 sin(ωt − δ) where δ = arctan(ωL/R) R + ω L2 Nguyen Thieu Huy (HUST) Infinite Series and Diff Eq 9/9 Hence, I (t) = C e −αt + E0 L α sin ωt − ω cos ωt α2 + ω = E0 sin(ωt − δ) where δ = arctan(ωL/R) R + ω L2 The exponential term will approach zero as t → ∞ This mean that after some time the current I (t) will execute practically harmonic oscillations C e −αt + √ Nguyen Thieu Huy (HUST) Infinite Series and Diff Eq 9/9 Hence, I (t) = C e −αt + E0 L α sin ωt − ω cos ωt α2 + ω = E0 sin(ωt − δ) where δ = arctan(ωL/R) R + ω L2 The exponential term will approach zero as t → ∞ This mean that after some time the current I (t) will execute practically harmonic oscillations C e −αt + √ Nguyen Thieu Huy (HUST) Infinite Series and Diff Eq 9/9 Hence, I (t) = C e −αt + E0 L α sin ωt − ω cos ωt α2 + ω = E0 sin(ωt − δ) where δ = arctan(ωL/R) R + ω L2 The exponential term will approach zero as t → ∞ This mean that after some time the current I (t) will execute practically harmonic oscillations C e −αt + √ Existence and Uniqueness Theorem ∂f Let the functions f (x, y ) and ∂y be continuous in some rectangle α < x < β, γ < y < δ containing the point (x0 , y0 ) Then, in some interval (x0 − h, x0 + h) contained in (α, β), there is a unique solution y = ϕ(x), x ∈ (x0 − h, x0 + h), of the initial value problem y = f (x, y ), y (x0 ) = y0 Nguyen Thieu Huy (HUST) Infinite Series and Diff Eq 9/9 ... the (differential) equations Analysis of the Model (A) To solve equations (B) To find the properties of solutions (C) If it is too difficult to solve the equations, try to approximate the solutions. .. the (differential) equations Analysis of the Model (A) To solve equations (B) To find the properties of solutions (C) If it is too difficult to solve the equations, try to approximate the solutions. .. The Eq ⇔ xy2 dx − x1 dy = which is exact The solution is Linear Equations dx x y x = x2 = C 4.1 Definition A first- order linear differential equation has the form y + p(x)y = q(x) v dI R E Example