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- 1. Chapter 1 First order differential equations Lecture3 ODE (Lecture3) Introduction to ODE-Math214 Dr Lamia Jaafar 1 / 22
- 2. Linear equations Follow-up ODE (Lecture3) Introduction to ODE-Math214 Dr Lamia Jaafar 2 / 22
- 3. Example Let the ODE (E) deﬁned by (x 2 − 9)y + xy = x. 1 Give the general solution of (E). 2 Precise a domain over which the solution is well deﬁned. 3 Discuss existence and uniqueness of the solution if we add the 1 initial value condition y(5) = . 4 Solution Step1:The general solution of (E) is given by y = yh + yp , where yh is the solution of the homogeneous equation associated to (E) and (yp ) is a particular solution of (E). ODE (Lecture3) Introduction to ODE-Math214 Dr Lamia Jaafar 3 / 22
- 4. Example Let the ODE (E) deﬁned by (x 2 − 9)y + xy = x. 1 Give the general solution of (E). 2 Precise a domain over which the solution is well deﬁned. 3 Discuss existence and uniqueness of the solution if we add the 1 initial value condition y(5) = . 4 Solution Step1:The general solution of (E) is given by y = yh + yp , where yh is the solution of the homogeneous equation associated to (E) and (yp ) is a particular solution of (E). ODE (Lecture3) Introduction to ODE-Math214 Dr Lamia Jaafar 3 / 22
- 5. Step2: the homogeneous solution yh is determined by solving the separate variables equation (x 2 − 9)y + xy = 0 ⇔ then yh (x) = 1 x dy = − 2 dx, x = ±3 y x −9 λ |x 2 − 9| , x = ±3, λ ∈ R. Step3: a particular solution of (E) can be determined by taking yp = 1. We deduce that the general solution of (E) is given by λ + 1, x = ±3, λ ∈ R. y (x) = 2 − 9| |x The points 3 and −3 are said to be singular points. ODE (Lecture3) Introduction to ODE-Math214 Dr Lamia Jaafar 4 / 22
- 6. Step2: the homogeneous solution yh is determined by solving the separate variables equation (x 2 − 9)y + xy = 0 ⇔ then yh (x) = 1 x dy = − 2 dx, x = ±3 y x −9 λ |x 2 − 9| , x = ±3, λ ∈ R. Step3: a particular solution of (E) can be determined by taking yp = 1. We deduce that the general solution of (E) is given by λ + 1, x = ±3, λ ∈ R. y (x) = 2 − 9| |x The points 3 and −3 are said to be singular points. ODE (Lecture3) Introduction to ODE-Math214 Dr Lamia Jaafar 4 / 22
- 7. Step2: the homogeneous solution yh is determined by solving the separate variables equation (x 2 − 9)y + xy = 0 ⇔ then yh (x) = 1 x dy = − 2 dx, x = ±3 y x −9 λ |x 2 − 9| , x = ±3, λ ∈ R. Step3: a particular solution of (E) can be determined by taking yp = 1. We deduce that the general solution of (E) is given by λ + 1, x = ±3, λ ∈ R. y (x) = 2 − 9| |x The points 3 and −3 are said to be singular points. ODE (Lecture3) Introduction to ODE-Math214 Dr Lamia Jaafar 4 / 22
- 8. If x ∈] − ∞, −3[, the general solution is given by λ y1 (x) = √ + 1. x2 − 9 If x ∈] − 3, 3[, the general solution is given by λ + 1. y2 (x) = √ 9 − x2 If x ∈]3, +∞[, the general solution is given by λ + 1. y3 (x) = √ 2−9 x 1 , we solve the equation on (3, +∞), on which the 4 theorem of existence and uniqueness are assured: By taking y (5) = ∃!y(x) = √ ODE (Lecture3) 4 x2 −9 + 1, x ∈ (3, +∞). Introduction to ODE-Math214 Dr Lamia Jaafar 5 / 22
- 9. If x ∈] − ∞, −3[, the general solution is given by λ y1 (x) = √ + 1. x2 − 9 If x ∈] − 3, 3[, the general solution is given by λ + 1. y2 (x) = √ 9 − x2 If x ∈]3, +∞[, the general solution is given by λ + 1. y3 (x) = √ 2−9 x 1 , we solve the equation on (3, +∞), on which the 4 theorem of existence and uniqueness are assured: By taking y (5) = ∃!y(x) = √ ODE (Lecture3) 4 x2 −9 + 1, x ∈ (3, +∞). Introduction to ODE-Math214 Dr Lamia Jaafar 5 / 22
- 10. If x ∈] − ∞, −3[, the general solution is given by λ y1 (x) = √ + 1. x2 − 9 If x ∈] − 3, 3[, the general solution is given by λ + 1. y2 (x) = √ 9 − x2 If x ∈]3, +∞[, the general solution is given by λ + 1. y3 (x) = √ 2−9 x 1 , we solve the equation on (3, +∞), on which the 4 theorem of existence and uniqueness are assured: By taking y (5) = ∃!y(x) = √ ODE (Lecture3) 4 x2 −9 + 1, x ∈ (3, +∞). Introduction to ODE-Math214 Dr Lamia Jaafar 5 / 22
- 11. Example Let (E) deﬁned by xy − 4y = x 5 ex . 1 2 3 Verify that yp (x) = x 4 ex veriﬁes the equation (E). Find the general solution y of (E) and precise a domain over which the solution is well deﬁned. Discuss the existence and uniqueness of the IVP if the following initial conditions are given y(0) = 0, y(0) = y0 , y0 > 0, y(x0 ) = y0 , x0 > 0, y0 > 0. ODE (Lecture3) Introduction to ODE-Math214 Dr Lamia Jaafar 6 / 22
- 12. 1 2 3 It is clear that y (x) = x 4 ex veriﬁes (E). The general solution of (E) is given by y = yh + yp , where yh is the solution of the homogeneous equation and yp is a particular solution of (E). For x = 0, we have xy − 4y = 0 ⇔ y = λx 4 , λ ∈ R, then for x = 0, y(x) = λx 4 + x 4 ex . We conclude that I = (0, ∞) is a domain over which y is well deﬁned. We have y(0) = 0 ⇒ a non uniqueness of the solution. y(0) = y0 , y0 > 0 ⇒ a non existence of the solution. y(x0 ) = y0 , x0 > 0, y0 > 0 ⇒ existence and uniqueness of the solution. ODE (Lecture3) Introduction to ODE-Math214 Dr Lamia Jaafar 7 / 22
- 13. 1 2 3 It is clear that y (x) = x 4 ex veriﬁes (E). The general solution of (E) is given by y = yh + yp , where yh is the solution of the homogeneous equation and yp is a particular solution of (E). For x = 0, we have xy − 4y = 0 ⇔ y = λx 4 , λ ∈ R, then for x = 0, y(x) = λx 4 + x 4 ex . We conclude that I = (0, ∞) is a domain over which y is well deﬁned. We have y(0) = 0 ⇒ a non uniqueness of the solution. y(0) = y0 , y0 > 0 ⇒ a non existence of the solution. y(x0 ) = y0 , x0 > 0, y0 > 0 ⇒ existence and uniqueness of the solution. ODE (Lecture3) Introduction to ODE-Math214 Dr Lamia Jaafar 7 / 22
- 14. 1 2 3 It is clear that y (x) = x 4 ex veriﬁes (E). The general solution of (E) is given by y = yh + yp , where yh is the solution of the homogeneous equation and yp is a particular solution of (E). For x = 0, we have xy − 4y = 0 ⇔ y = λx 4 , λ ∈ R, then for x = 0, y(x) = λx 4 + x 4 ex . We conclude that I = (0, ∞) is a domain over which y is well deﬁned. We have y(0) = 0 ⇒ a non uniqueness of the solution. y(0) = y0 , y0 > 0 ⇒ a non existence of the solution. y(x0 ) = y0 , x0 > 0, y0 > 0 ⇒ existence and uniqueness of the solution. ODE (Lecture3) Introduction to ODE-Math214 Dr Lamia Jaafar 7 / 22
- 15. Method of variation of parameters Variation of parameters (or variation of constants) is a general method for determining the general solution of a linear ODE (E) : a(x)y (x) + b(x)y (x) = φ(x). Let (Eh ) : a(x)y (x) + b(x)y (x) = 0, then −b(x) dx, λ ∈ R. a(x) Let us suppose that λ is a function depending of the variable x, i.e y (x) = λ(x) exp(f (x)), then yh (x) = λef (x) , f (x) = (E) ⇔ λ(x) = We obtain λ(x) = ODE (Lecture3) c(x) exp(−f (x)). a(x) c(x) exp(−f (x))dx + C, C ∈ R. a(x) Introduction to ODE-Math214 Dr Lamia Jaafar 8 / 22
- 16. Method of variation of parameters Variation of parameters (or variation of constants) is a general method for determining the general solution of a linear ODE (E) : a(x)y (x) + b(x)y (x) = φ(x). Let (Eh ) : a(x)y (x) + b(x)y(x) = 0, then −b(x) dx, λ ∈ R. a(x) Let us suppose that λ is a function depending of the variable x, i.e y (x) = λ(x) exp(f (x)), then yh (x) = λef (x) , f (x) = (E) ⇔ λ(x) = We obtain λ(x) = ODE (Lecture3) c(x) exp(−f (x)). a(x) c(x) exp(−f (x))dx + C, C ∈ R. a(x) Introduction to ODE-Math214 Dr Lamia Jaafar 8 / 22
- 17. Method of variation of parameters Variation of parameters (or variation of constants) is a general method for determining the general solution of a linear ODE (E) : a(x)y (x) + b(x)y (x) = φ(x). Let (Eh ) : a(x)y (x) + b(x)y(x) = 0, then −b(x) dx, λ ∈ R. a(x) Let us suppose that λ is a function depending of the variable x, i.e y (x) = λ(x) exp(f (x)), then yh (x) = λef (x) , f (x) = (E) ⇔ λ(x) = We obtain λ(x) = ODE (Lecture3) c(x) exp(−f (x)). a(x) c(x) exp(−f (x))dx + C, C ∈ R. a(x) Introduction to ODE-Math214 Dr Lamia Jaafar 8 / 22
- 18. Method of variation of parameters Variation of parameters (or variation of constants) is a general method for determining the general solution of a linear ODE (E) : a(x)y (x) + b(x)y (x) = φ(x). Let (Eh ) : a(x)y (x) + b(x)y(x) = 0, then −b(x) dx, λ ∈ R. a(x) Let us suppose that λ is a function depending of the variable x, i.e y (x) = λ(x) exp(f (x)), then yh (x) = λef (x) , f (x) = (E) ⇔ λ(x) = We obtain λ(x) = ODE (Lecture3) c(x) exp(−f (x)). a(x) c(x) exp(−f (x))dx + C, C ∈ R. a(x) Introduction to ODE-Math214 Dr Lamia Jaafar 8 / 22
- 19. Conclusion We conclude that the general solution of a linear ordinary equation is given by y(x) = C exp(f (x)) + exp(f (x)) = ↓ yh (x) + c(x) exp(−f (x))dx a(x) ↓ yp (x) Examples Solve the IVP y +y =x y (0) = 4 and precise the domain of existence and uniqueness of the solution. ODE (Lecture3) Introduction to ODE-Math214 Dr Lamia Jaafar 9 / 22
- 20. Conclusion We conclude that the general solution of a linear ordinary equation is given by y(x) = C exp(f (x)) + exp(f (x)) = ↓ yh (x) + c(x) exp(−f (x))dx a(x) ↓ yp (x) Examples Solve the IVP y +y =x y (0) = 4 and precise the domain of existence and uniqueness of the solution. ODE (Lecture3) Introduction to ODE-Math214 Dr Lamia Jaafar 9 / 22
- 21. Exact equations ODE (Lecture3) Introduction to ODE-Math214 Dr Lamia Jaafar 10 / 22
- 22. Deﬁnition an exact differential equation is an ODE of ﬁrst order which can be expressed as follows M(x, y)dx + N(x, y)dy = 0, (1) where M and N are functions in both the variables x and y, such that ∂M ∂N = . ∂y ∂x Remark Let us suppose that f is a given real function deﬁned in both variables ∂f ∂f x, y , then its differential is given by df = dx + dy . ∂x ∂y ∂f ∂f If f is constant we obtain the equation dx + dy = 0, then the ∂x ∂y equation (1) corresponds to a differential of f (x, y) = Cst. ODE (Lecture3) Introduction to ODE-Math214 Dr Lamia Jaafar 11 / 22
- 23. Deﬁnition an exact differential equation is an ODE of ﬁrst order which can be expressed as follows M(x, y)dx + N(x, y)dy = 0, (1) where M and N are functions in both the variables x and y, such that ∂M ∂N = . ∂y ∂x Remark Let us suppose that f is a given real function deﬁned in both variables ∂f ∂f x, y , then its differential is given by df = dx + dy . ∂x ∂y ∂f ∂f If f is constant we obtain the equation dx + dy = 0, then the ∂x ∂y equation (1) corresponds to a differential of f (x, y) = Cst. ODE (Lecture3) Introduction to ODE-Math214 Dr Lamia Jaafar 11 / 22
- 24. Example x 2 y 3 dx + x 3 y 2 dy = 0, is an exact equation such that f (x, y) = Then the solution is given by an explicit one deﬁned as 1 3 3 x y = C, C ∈ R. 3 1 3 3 x y . 3 Examples 2xydx + (x 2 − 1)dy = 0, 1 y y(1 − x 2 ) − xy 2 + sin 2x, 2 are exact equations too, but f (x, y) cannot be easily deduced. ODE (Lecture3) Introduction to ODE-Math214 Dr Lamia Jaafar 12 / 22
- 25. Example x 2 y 3 dx + x 3 y 2 dy = 0, is an exact equation such that f (x, y) = Then the solution is given by an explicit one deﬁned as 1 3 3 x y = C, C ∈ R. 3 1 3 3 x y . 3 Examples 2xydx + (x 2 − 1)dy = 0, 1 y y(1 − x 2 ) − xy 2 + sin 2x, 2 are exact equations too, but f (x, y) cannot be easily deduced. ODE (Lecture3) Introduction to ODE-Math214 Dr Lamia Jaafar 12 / 22
- 26. Remark ∂f ∂f and N(x, y ) = , we deduce that to solve (1), we ∂x ∂y ∂f = M and to integrate with respect have to ﬁnd M (or N) such that ∂x ∂f to x (such that = N and to integrate with respect to y .) ∂y As M(x, y ) = Examples Solve the previous exact equations. ODE (Lecture3) Introduction to ODE-Math214 Dr Lamia Jaafar 13 / 22
- 27. Remark ∂f ∂f and N(x, y ) = , we deduce that to solve (1), we ∂x ∂y ∂f = M and to integrate with respect have to ﬁnd M (or N) such that ∂x ∂f to x (such that = N and to integrate with respect to y .) ∂y As M(x, y ) = Examples Solve the previous exact equations. ODE (Lecture3) Introduction to ODE-Math214 Dr Lamia Jaafar 13 / 22
- 28. Solution of (E) : 2xydx + (x 2 − 1)dy = 0 Let M(x, y) = 2xy and N(x, y) = x 2 − 1. ∂M ∂N As = then (E) is an exact equation. We deduce that it exists f ∂y ∂x deﬁned on a domain I such that ∂f 2 ∂x = 2xy, f (x, y) = x y + F (y), ⇒ ∂f ∂f 2−1 = x 2 − 1 = x 2 + F (y ) ∂y ∂y = x we deduce that F (y) = −1 ⇒ F (y ) = −y + Cst ⇒ f (x, y) = x 2 y − y + Cst, then an implicit solution is giving by x 2 y − y − C, C ∈ R and an explicit solution C is y = 2 , x = ±1. x −1 The solution of the initial ODE is deﬁned on any interval not containing either 1 or −1, i.e a domain of the solution y is ] − ∞, −1[ or ] − 1, 1[ or ]1, +∞[. ODE (Lecture3) Introduction to ODE-Math214 Dr Lamia Jaafar 14 / 22
- 29. Solution of (E) : 2xydx + (x 2 − 1)dy = 0 Let M(x, y) = 2xy and N(x, y) = x 2 − 1. ∂M ∂N As = then (E) is an exact equation. We deduce that it exists f ∂y ∂x deﬁned on a domain I such that ∂f 2 ∂x = 2xy, f (x, y) = x y + F (y), ⇒ ∂f ∂f 2−1 = x 2 − 1 = x 2 + F (y ) ∂y ∂y = x we deduce that F (y) = −1 ⇒ F (y ) = −y + Cst ⇒ f (x, y) = x 2 y − y + Cst, then an implicit solution is giving by x 2 y − y − C, C ∈ R and an explicit solution C is y = 2 , x = ±1. x −1 The solution of the initial ODE is deﬁned on any interval not containing either 1 or −1, i.e a domain of the solution y is ] − ∞, −1[ or ] − 1, 1[ or ]1, +∞[. ODE (Lecture3) Introduction to ODE-Math214 Dr Lamia Jaafar 14 / 22
- 30. Integrating factor Remark Let (E1 ) : M(x, y )dx + N(x, y )dy = 0 such that ∂N ∂M = , then (E1 ) is ∂y ∂x not an exact equation. Question How we can transform it to an exact one? The method proposed Find µ(x, y) such that (E2 ) : µ(x, y)M(x, y )dx + µ(x, y)N(x, y)dy = 0 is an exact solution. We obtain ∂ ∂ ∂µ ∂µ ∂M ∂N (µM) = (µN) ⇒ N− M = µ( − ). ∂y ∂x ∂x ∂y ∂y ∂x Two cases are then possible to deduce an integrating factor µ. ODE (Lecture3) Introduction to ODE-Math214 Dr Lamia Jaafar 15 / 22
- 31. Integrating factor Remark Let (E1 ) : M(x, y )dx + N(x, y )dy = 0 such that ∂N ∂M = , then (E1 ) is ∂y ∂x not an exact equation. Question How we can transform it to an exact one? The method proposed Find µ(x, y) such that (E2 ) : µ(x, y)M(x, y )dx + µ(x, y)N(x, y)dy = 0 is an exact solution. We obtain ∂ ∂ ∂µ ∂µ ∂M ∂N (µM) = (µN) ⇒ N− M = µ( − ). ∂y ∂x ∂x ∂y ∂y ∂x Two cases are then possible to deduce an integrating factor µ. ODE (Lecture3) Introduction to ODE-Math214 Dr Lamia Jaafar 15 / 22
- 32. Integrating factor Remark Let (E1 ) : M(x, y )dx + N(x, y )dy = 0 such that ∂N ∂M = , then (E1 ) is ∂y ∂x not an exact equation. Question How we can transform it to an exact one? The method proposed Find µ(x, y) such that (E2 ) : µ(x, y)M(x, y )dx + µ(x, y)N(x, y)dy = 0 is an exact solution. We obtain ∂ ∂ ∂µ ∂µ ∂M ∂N (µM) = (µN) ⇒ N− M = µ( − ). ∂y ∂x ∂x ∂y ∂y ∂x Two cases are then possible to deduce an integrating factor µ. ODE (Lecture3) Introduction to ODE-Math214 Dr Lamia Jaafar 15 / 22
- 33. Integrating factor Remark Let (E1 ) : M(x, y )dx + N(x, y )dy = 0 such that ∂N ∂M = , then (E1 ) is ∂y ∂x not an exact equation. Question How we can transform it to an exact one? The method proposed Find µ(x, y) such that (E2 ) : µ(x, y)M(x, y )dx + µ(x, y)N(x, y)dy = 0 is an exact solution. We obtain ∂ ∂ ∂µ ∂µ ∂M ∂N (µM) = (µN) ⇒ N− M = µ( − ). ∂y ∂x ∂x ∂y ∂y ∂x Two cases are then possible to deduce an integrating factor µ. ODE (Lecture3) Introduction to ODE-Math214 Dr Lamia Jaafar 15 / 22
- 34. Let µx = ∂µ , µy ∂x = ∂µ , Mx ∂y = ∂M , My ∂x = ∂M , Nx ∂y = ∂N , Ny ∂x = ∂N . ∂y My − Nx depends of x alone then we take N My − Nx dx , µ(x) = exp N Nx − My If depends of y alone then we take M Nx − My µ(y) = exp dy . M If Remark If we consider a linear equation a(x)y (x) + b(x)y (x) = φ(x) equivalent b(x) φ(x) to y (x) + P(x)y(x) = ψ(x), where P(x) = and ψ(x) = . a(x) a(x) Then exp − P(x)dx is also called an integrating factor. ODE (Lecture3) Introduction to ODE-Math214 Dr Lamia Jaafar 16 / 22
- 35. Let µx = ∂µ , µy ∂x = ∂µ , Mx ∂y = ∂M , My ∂x = ∂M , Nx ∂y = ∂N , Ny ∂x = ∂N . ∂y My − Nx depends of x alone then we take N My − Nx dx , µ(x) = exp N Nx − My If depends of y alone then we take M Nx − My µ(y) = exp dy . M If Remark If we consider a linear equation a(x)y (x) + b(x)y (x) = φ(x) equivalent b(x) φ(x) to y (x) + P(x)y(x) = ψ(x), where P(x) = and ψ(x) = . a(x) a(x) Then exp − P(x)dx is also called an integrating factor. ODE (Lecture3) Introduction to ODE-Math214 Dr Lamia Jaafar 16 / 22
- 36. Example 1 2 3 Is (E) : xydx + (2x 2 + 3y 2 − 20)dy = 0 an exact equation? Prove that an implicit solution of (E) is giving by 1 2 4 1 6 x y + y − 5y 4 = C, C ∈ R. 2 2 Solve the initial value problem associated to (E) if y(1) = 1. ODE (Lecture3) Introduction to ODE-Math214 Dr Lamia Jaafar 17 / 22
- 37. Solutions by substitutions ODE (Lecture3) Introduction to ODE-Math214 Dr Lamia Jaafar 18 / 22
- 38. Deﬁnition A function f is said to be homogeneous of degree α if it veriﬁes f (tx, ty) = t α f (x, y ), α ∈ R. Examples f deﬁned by f (x, y) = x 3 + y 3 is homogeneous of degree 3. g given by g(x, y ) = x 2 + y 2 + xy is homogeneous of degree 2. h verifying h(x, y) = x + y + xy is not homogeneous. Remark If both M and N are homogeneous functions of degree α, then the ﬁrst order differential equation (E) : M(x, y)dx + N(x, y)dy = 0 is said to be homogeneous of degree α. Example The ODE (x 2 + y 2 )dx + (x 2 − xy)dy = 0 is an homogeneous ﬁrst order differential equation of degree 2. ODE (Lecture3) Introduction to ODE-Math214 Dr Lamia Jaafar 19 / 22
- 39. Deﬁnition A function f is said to be homogeneous of degree α if it veriﬁes f (tx, ty) = t α f (x, y ), α ∈ R. Examples f deﬁned by f (x, y) = x 3 + y 3 is homogeneous of degree 3. g given by g(x, y ) = x 2 + y 2 + xy is homogeneous of degree 2. h verifying h(x, y) = x + y + xy is not homogeneous. Remark If both M and N are homogeneous functions of degree α, then the ﬁrst order differential equation (E) : M(x, y)dx + N(x, y)dy = 0 is said to be homogeneous of degree α. Example The ODE (x 2 + y 2 )dx + (x 2 − xy)dy = 0 is an homogeneous ﬁrst order differential equation of degree 2. ODE (Lecture3) Introduction to ODE-Math214 Dr Lamia Jaafar 19 / 22
- 40. Deﬁnition A function f is said to be homogeneous of degree α if it veriﬁes f (tx, ty) = t α f (x, y ), α ∈ R. Examples f deﬁned by f (x, y) = x 3 + y 3 is homogeneous of degree 3. g given by g(x, y ) = x 2 + y 2 + xy is homogeneous of degree 2. h verifying h(x, y) = x + y + xy is not homogeneous. Remark If both M and N are homogeneous functions of degree α, then the ﬁrst order differential equation (E) : M(x, y)dx + N(x, y)dy = 0 is said to be homogeneous of degree α. Example The ODE (x 2 + y 2 )dx + (x 2 − xy)dy = 0 is an homogeneous ﬁrst order differential equation of degree 2. ODE (Lecture3) Introduction to ODE-Math214 Dr Lamia Jaafar 19 / 22
- 41. Deﬁnition A function f is said to be homogeneous of degree α if it veriﬁes f (tx, ty) = t α f (x, y ), α ∈ R. Examples f deﬁned by f (x, y) = x 3 + y 3 is homogeneous of degree 3. g given by g(x, y ) = x 2 + y 2 + xy is homogeneous of degree 2. h verifying h(x, y) = x + y + xy is not homogeneous. Remark If both M and N are homogeneous functions of degree α, then the ﬁrst order differential equation (E) : M(x, y)dx + N(x, y)dy = 0 is said to be homogeneous of degree α. Example The ODE (x 2 + y 2 )dx + (x 2 − xy)dy = 0 is an homogeneous ﬁrst order differential equation of degree 2. ODE (Lecture3) Introduction to ODE-Math214 Dr Lamia Jaafar 19 / 22
- 42. Method of solving homogeneous ODE of degree α. Let y = ux then the equation (E) gives M(1, u)dx + N(1, u)dy = 0. Or dy = udx + xdu, then we obtain a separable variables equation in x and u given by N(1, u)du dx =− du. x M(1, u) + uN(1, u) Example Prove that an implicit solution of (x 2 + y 2 )dx + (x 2 − xy)dy = 0, is y giving by (x + y)2 = λx exp( ), λ ∈ R. x ODE (Lecture3) Introduction to ODE-Math214 Dr Lamia Jaafar 20 / 22
- 43. Method of solving homogeneous ODE of degree α. Let y = ux then the equation (E) gives M(1, u)dx + N(1, u)dy = 0. Or dy = udx + xdu, then we obtain a separable variables equation in x and u given by N(1, u)du dx =− du. x M(1, u) + uN(1, u) Example Prove that an implicit solution of (x 2 + y 2 )dx + (x 2 − xy)dy = 0, is y giving by (x + y)2 = λx exp( ), λ ∈ R. x ODE (Lecture3) Introduction to ODE-Math214 Dr Lamia Jaafar 20 / 22
- 44. Some particular equations Bernouilli equation The Bernouilli equation is a ﬁrst order differential equation of the form y (x) + a(x)y (x) = b(x)y α , α ∈ R. Remarks 1 If α = 0 or α = 1 then we obtain a linear equation. 2 If α = 0 and α = 1, the solution can be deduced by taking u = y 1−α . Example Solve the ODE xy + y = y 2 ln x for x > 0. ODE (Lecture3) Introduction to ODE-Math214 Dr Lamia Jaafar 21 / 22
- 45. Some particular equations Bernouilli equation The Bernouilli equation is a ﬁrst order differential equation of the form y (x) + a(x)y (x) = b(x)y α , α ∈ R. Remarks 1 If α = 0 or α = 1 then we obtain a linear equation. 2 If α = 0 and α = 1, the solution can be deduced by taking u = y 1−α . Example Solve the ODE xy + y = y 2 ln x for x > 0. ODE (Lecture3) Introduction to ODE-Math214 Dr Lamia Jaafar 21 / 22
- 46. Some particular equations Bernouilli equation The Bernouilli equation is a ﬁrst order differential equation of the form y (x) + a(x)y (x) = b(x)y α , α ∈ R. Remarks 1 If α = 0 or α = 1 then we obtain a linear equation. 2 If α = 0 and α = 1, the solution can be deduced by taking u = y 1−α . Example Solve the ODE xy + y = y 2 ln x for x > 0. ODE (Lecture3) Introduction to ODE-Math214 Dr Lamia Jaafar 21 / 22
- 47. Ricatti equation The Ricatti equation is a ﬁrst order differential equation of the form y (x) + a(x)y (x) = b(x)y 2 + c(x). Remarks 1 If c = 0 then we obtain a Bernouilli equation, with α = 2. 2 If c = 0, the solution of the ricatti equation can be deduced by taking z(x) = y (x) − yp (x), where yp is a particular solution of the initial Ricatti equation. Example Let (E) : x 3 y + y 2 + yx 2 + 2x 4 = 0. 1 Verify that yp = −x 2 is a particular solution of (E). 2 Find the general solution of (E). ODE (Lecture3) Introduction to ODE-Math214 Dr Lamia Jaafar 22 / 22
- 48. Ricatti equation The Ricatti equation is a ﬁrst order differential equation of the form y (x) + a(x)y (x) = b(x)y 2 + c(x). Remarks 1 If c = 0 then we obtain a Bernouilli equation, with α = 2. 2 If c = 0, the solution of the ricatti equation can be deduced by taking z(x) = y (x) − yp (x), where yp is a particular solution of the initial Ricatti equation. Example Let (E) : x 3 y + y 2 + yx 2 + 2x 4 = 0. 1 Verify that yp = −x 2 is a particular solution of (E). 2 Find the general solution of (E). ODE (Lecture3) Introduction to ODE-Math214 Dr Lamia Jaafar 22 / 22
- 49. Ricatti equation The Ricatti equation is a ﬁrst order differential equation of the form y (x) + a(x)y (x) = b(x)y 2 + c(x). Remarks 1 If c = 0 then we obtain a Bernouilli equation, with α = 2. 2 If c = 0, the solution of the ricatti equation can be deduced by taking z(x) = y (x) − yp (x), where yp is a particular solution of the initial Ricatti equation. Example Let (E) : x 3 y + y 2 + yx 2 + 2x 4 = 0. 1 Verify that yp = −x 2 is a particular solution of (E). 2 Find the general solution of (E). ODE (Lecture3) Introduction to ODE-Math214 Dr Lamia Jaafar 22 / 22

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