Chapter 2: First-Order Differential Equations Part 2 Department of - PowerPoint PPT Presentation

Exact Equations Solutions by Substitutions Summary Chapter 2: First-Order Differential Equations – Part 2 Department of Electrical Engineering National Taiwan University ihwang@ntu.edu.tw October 3, 2013 DE Lecture 4 王奕翔 王奕翔

Exact Equations We will not follow the order in the textbook. Instead, Solutions by Substitutions DE Lecture 4 Organization of Lectures in Chapter 2 and 3 Summary ��� Separable DE (2-2) ��� �� DE Linear (2-1) (2-3) Models (3-1) ���� Exact DE Nonlinear (2-6) (2-4) Models (3-2) ���� (2-5) 王奕翔

Exact Equations Solutions by Substitutions Summary 1 Exact Equations 2 Solutions by Substitutions Homogeneous Equations Bernoulli’s Equation 3 Summary DE Lecture 4 王奕翔

Exact Equations Solutions by Substitutions Summary DE Lecture 4 今天，我要出一道微分方程的考題，我可以從哪邊下手？ 王奕翔

Exact Equations Solutions by Substitutions DE Lecture 4 Summary One proposal: reverse engineering – 先寫下解答，再反推回去方程式 1 Set up the solution curve: G ( x , y ) = 0 (can be an implicit solution) and an initial point ( x 0 , y 0 ) . 2 Compute the differential of G ( x , y ) : d ( G ( x , y )) = ∂ G ∂ x dx + ∂ G ∂ y dy 3 Since G ( x , y ) = 0 , we have 0 = d ( G ( x , y )) = ∂ G ∂ x dx + ∂ G ∂ y dy 4 Let ∂ G ∂ x = M ( x , y ) and ∂ G ∂ y = N ( x , y ) . Then, we have a DE: dx = − M ( x , y ) M ( x , y ) dx + N ( x , y ) dy = 0 = ⇒ dy N ( x , y ) 王奕翔

Exact Equations Solutions by Substitutions the above observation. We shall develop a general method of solving this kind of DE based on DE Lecture 4 Summary 解題者觀點：看到一個一階常微分方程，若能將其化為 ∂ F ∂ x dx + ∂ F ∂ y dy = 0 We can get the solution: F ( x , y ) = c , where c = F ( x 0 , y 0 ) . Note : the function F ( x , y ) you get may not be the same as the designer’s choice G ( x , y ) . Because the designer chose G ( x , y ) = 0 as his/her solution, while what you get is F ( x , y ) = F ( x 0 , y 0 ) . Nevertheless, G ( x , y ) = F ( x , y ) − F ( x 0 , y 0 ) . 王奕翔

Exact Equations Exact Equation if the LHS is an exact differential. . Solutions by Substitutions Hint: Question: How to check if a differential expression is an exact differential? DE Lecture 4 Exact Differential and Exact Equation Definition (Exact Equation) Summary A differential expression M ( x , y ) dx + N ( x , y ) dy is an Exact Differential if it is the differential of some function z = F ( x , y ) , that is, dz = M ( x , y ) dx + N ( x , y ) dy . A first-order DE of the form M ( x , y ) dx + N ( x , y ) dy = 0 is said to be an ( ) ( ∂ F ) ∂ ∂ ∂ F = ∂ y ∂ x ∂ x ∂ y 王奕翔

Exact Equations Proof. procedure later. In fact, this is the procedure of solving an exact DE. We will outline the . Solutions by Substitutions DE Lecture 4 derivatives. Then, Summary Criterion for an Exact Differential Theorem Let M ( x , y ) and N ( x , y ) be continuous and have continuous first partial ⇒ ∂ M ∂ y = ∂ N M ( x , y ) dx + N ( x , y ) dy is an exact differential ⇐ ∂ x ( ) ( ∂ F ) ∂ ∂ ∂ F “ ⇒ ”: Simply because = ∂ y ∂ x ∂ x ∂ y “ ⇐ ”: We just need to construct a function z = F ( x , y ) such that dz = M ( x , y ) dx + N ( x , y ) dy . 王奕翔

Exact Equations Check if the DE is exact: respect to x ? Solutions by Substitutions DE Lecture 4 Solve Summary Solving an Exact DE Example ( ) ( ) e 2 y − y cos ( xy ) 2 xe 2 y − x cos ( xy ) + 2 y dx + dy = 0 . A: Let M ( x , y ) = e 2 y − y cos ( xy ) and N ( x , y ) = 2 xe 2 y − x cos ( xy ) + 2 y . ( ) ∂ M ∂ y = ∂ e 2 y − y cos ( xy ) = 2 e 2 y − cos ( xy ) + xy sin ( xy ) ∂ y ( ) ∂ N ∂ x = ∂ 2 xe 2 y − x cos ( xy ) + 2 y = 2 e 2 y − cos ( xy ) + xy sin ( xy ) ∂ x Since M = ∂ F ∂ x and we want to find F , why not integrate M with ∫ { } e 2 y − y cos ( xy ) dx + g ( y ) = e 2 y x − sin ( xy ) + g ( y ) . F ( x , y ) = 王奕翔

Exact Equations Solve Solutions by Substitutions be determined. Example Summary Solving an Exact DE DE Lecture 4 ( ) ( ) e 2 y − y cos ( xy ) 2 xe 2 y − x cos ( xy ) + 2 y dx + dy = 0 . A: Let M ( x , y ) = e 2 y − y cos ( xy ) and N ( x , y ) = 2 xe 2 y − x cos ( xy ) + 2 y . So far we found that F ( x , y ) = e 2 y x − sin ( xy ) + g ( y ) where g ( y ) is yet to To find g ( y ) , we use the fact that N = ∂ F ∂ y : ( ) 2 xe 2 y − x cos ( xy ) + 2 y = ∂ F ∂ y = ∂ e 2 y x − sin ( xy ) + g ( y ) ∂ y = 2 xe 2 y − x cos ( xy ) + g ′ ( y ) = ⇒ g ( y ) = y 2 ⇒ dg dy = 2 y = Hence, F ( x , y ) = xe 2 y − sin ( xy ) + y 2 , and the implicit solution is xe 2 y − sin ( xy ) + y 2 = c . 王奕翔

Exact Equations 4 Take partial derivative with respect to y (or x ): dx Ndy = dy Mdx = Ndy Solutions by Substitutions Mdx DE Lecture 4 3 Integrate M with respect to x (or N with respect to y ): Summary General Procedure of Solving an DE Solving an Exact DE M ( x , y ) dx + N ( x , y ) dy = 0 Goal: Find z = F ( x , y ) such that dz = M ( x , y ) dx + N ( x , y ) dy = 0 . 1 Transform DE into the differential form: M ( x , y ) dx + N ( x , y ) dy = 0 . 2 Verify if it is exact: ∂ M = ∂ N ? ∂ y ∂ x ∫ ∫ F ( x , y ) = Mdx + g ( y ) (or F ( x , y ) = Ndy + h ( x ) ) ∂ F ∂ y = ∂ (∫ ) ∂ F ∂ x = ∂ (∫ ) + g ′ ( y ) = N ( x , y ) + h ′ ( x ) = M ( x , y ) ∂ y ∂ x ∫ ( N − ∂ ∫ ) ∫ ( M − ∂ ∫ ) ⇒ g ( y ) = ⇒ h ( x ) = ∂ y ∂ x 王奕翔

Exact Equations A: Check if this equation is exact: Solutions by Substitutions xy dy = DE Lecture 4 Example Summary Nonexact DE Made Exact dx = 20 xy − 2 y x − 3 x y , y (1) = − 1 Solve dy y = 20 − 2 y 2 − 3 x 2 dx = 20 xy − 2 y x − 3 x M ( x , y ) N ( x , y ) � �� � ���� (3 x 2 + 2 y 2 − 20) dx + ⇒ ( xy ) dy = 0 ∂ M ∂ y = 4 y ̸ = ∂ N ∂ x = y . Can we make it exact, by multiplying both M and N with some µ ( x , y ) ? 王奕翔

Exact Equations y This is a PDE ?! How to solve it? Solutions by Substitutions = DE Lecture 4 Nonexact DE Made Exact Example Summary M ( x , y ) N ( x , y ) � �� � ���� dx = 20 xy − 2 y x − 3 x (3 x 2 + 2 y 2 − 20) dx + ⇒ ( xy ) dy = 0 Solve dy Goal: find µ ( x , y ) such that ∂ ( µ M ) = ∂ ( µ N ) ∂ x . Let µ x := ∂µ ∂ x , µ y := ∂µ ∂ y ∂ y . ∂ ( µ M ) = µ y M + M y µ = (3 x 2 + 2 y 2 − 20) µ y + 4 y µ ∂ y ∂ ( µ N ) = µ x N + N x µ = ( xy ) µ x + y µ ∂ x ∂ ( µ M ) = ∂ ( µ N ) ⇒ (3 x 2 + 2 y 2 − 20) µ y + 4 y µ = ( xy ) µ x + y µ ⇐ ∂ y ∂ x 王奕翔

Exact Equations y = x Solutions by Substitutions = DE Lecture 4 Summary Example Nonexact DE Made Exact M ( x , y ) N ( x , y ) � �� � ���� dx = 20 xy − 2 y x − 3 x (3 x 2 + 2 y 2 − 20) dx + ⇒ ( xy ) dy = 0 Solve dy Focus on finding a function µ ( x , y ) such that (3 x 2 + 2 y 2 − 20) µ y + 4 y µ = ( xy ) µ x + y µ Let’s make some restriction: how about finding µ that only depends on x ? ⇒ xyd µ ⇒ d µ dx = 3 µ ⇒ µ = x 3 4 y µ = ( xy ) µ x + y µ = dx = 3 y µ = ( works! ) How about finding µ that only depends on y ? 3 y ⇒ d µ (3 x 2 + 2 y 2 − 20) µ y + 4 y µ = y µ = dy = − 3 x 2 + 2 y 2 − 20 µ ( still hard! ) 王奕翔

Exact Equations = = = dx = Solutions by Substitutions We then solve it by the procedures discussed before: DE Lecture 4 y Summary Nonexact DE Made Exact Example � � M ( x , y ) N ( x , y ) � �� � � �� � dx = 20 xy − 2 y x − 3 x x 3 (3 x 2 + 2 y 2 − 20) dx + ( x 4 y ) dy = 0 ⇒ Solve dy Finally we multiply both M ( x , y ) and N ( x , y ) with µ ( x ) = x 3 (see above). ∫ N = ∂ F Ndy = 1 2 x 4 y 2 + h ( x ) � � ⇒ F ( x , y ) = ∂ y = ( 1 ) M = ∂ F ⇒ x 3 (3 x 2 + 2 y 2 − 20) = ∂ dx = 2 x 3 y 2 + dh � 2 x 4 y 2 + dh ∂ x ∂ x ⇒ h ( x ) = 1 dx = 3 x 5 − 20 x 3 = 2 x 6 − 5 x 4 ⇒ dh ⇒ F ( x , y ) = 1 2 x 4 y 2 + 1 2 x 6 − 5 x 4 王奕翔

Exact Equations Solutions by Substitutions Exercise. Find an interval of definition for the above solution. = = To get an explicit solution, we see that DE Lecture 4 Example Summary Nonexact DE Made Exact dx = 20 xy − 2 y x − 3 x y , y (1) = − 1 Solve dy 2 x 4 y 2 + 1 2 x 6 − 5 x 4 = c . We arrive at an implicit solution: F ( x , y ) = 1 2 x 4 y 2 + 1 2 x 6 − 5 x 4 = c = − 4 . Plug in the initial condition, we get 1 2 x 4 y 2 + 1 1 2 x 6 − 5 x 4 = − 4 = ⇒ y 2 = 10 − x 2 − 8 x − 4 √ 10 − x 2 − 8 x − 4 ⇒ y = ± √ 10 − x 2 − 8 x − 4 ⇒ y = − 王奕翔

Exact Equations However in general this is a PDE which may be hard to solve. M = N Solutions by Substitutions = DE Lecture 4 Summary Nonexact DE M ( x , y ) dx + N ( x , y ) dy = 0 Made Exact Nonexact DE: M y − N x := ∆( x , y ) ̸ = 0 Key Idea 1 : Introduce a function µ ( x , y ) ( integrating factor ) to ensure ∂ ( µ M ) = ∂ ( µ N ) ⇐ ⇒ µ y M + µ M y = µ x N + µ N x ∂ y ∂ x Key Idea 2 : Restrict µ ( x , y ) to be µ ( x ) or µ ( y ) . Plan A: µ ( x , y ) = µ ( x ) = ⇒ µ y = 0 = ⇒ µ M y = µ x N + µ N x ⇒ d µ dx = M y − N x µ = ∆ N µ Plan B: µ ( x , y ) = µ ( y ) = ⇒ µ x = 0 = ⇒ µ y M + µ M y = µ N x dy = N x − M y µ = − ∆ ⇒ d µ M µ 王奕翔