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MA360Handout 6b

MA360Handout 6b - MA360 Sp2006 Kallfelz Handout 6b(week of...

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1 MA360 Handout 6b (week of Feb 20 ) Sp2006 Kallfelz This cursory review of linear ODEs (first and second order) is entirely example-driven. First-order Any garden variety first order linear ODE is of the form: ( ) ( ) ( ) ( ) ( ) t f t x t a t x t a = + 1 0 & , where: ( ) ( ) t x t x dt d & , for the unknown function x ( t ) and a 0 ( t ), a 1 ( t ), f ( t ) are given functions. We can always make this ODE simpler (by setting the leading term to be monic , i.e. the leading coefficient = 1) by dividing through by a 0 ( t ). 1 Hence: ( ) ( ) ( ) ( ) t g t x t p t x = + & , where: ( ) ( ) ( ) ( ) ( ) ( ) t a t f t a t a t g t p 0 0 1 , = = Lemma1: Multiplying (Eqn1) by the integrating factor ( ) ( ) = dt t p e t r makes it separable. Proof: Multiply both sides by r ( t ) and bringing all terms to one side: ( ) ( ) ( ) [ ] ( ) ( ) ( ) 0 = + - t x t r t r t g t x t p & If the above is separable, then the following condition must hold 2 : ( ) ( ) ( ) [ ] ( ) { } ( ) t r t r t g t x t p t x = - For the left hand side: ( ) ( ) ( ) [ ] ( ) { } ( ) ( ) ( ) ( ) ( ) ( ) ( ) t r t g t r t x t p t r t g t x t p x x x - = - Since p , g , r are all homogeneous functions depending on t only, they behave as constants, with respect to taking the partial derivative with respect to x . Hence the left side simplifies to: p ( t ) r ( t ) (Since: ( ) ( ) ( ) ( ) ( ) ( ) ( ) t r t p x t r t p t r t x t p x x = = and ( ) ( ) 0 = t r t g x ) On the other hand, the right hand side is just a total derivative with respect to t , since r is a homogeneous function with respect to t .: ( ) ( ) t r t r dt d t = So the above simplifies to: ( ) ( ) ( ) t r t r t p dt d = , which implies that ( ) ( ) = dt t p e t r . 1 Since the leading coefficient function must be nonzero. 2 This is true because if we assume there exists an exact solution curve φ ( x , t ) = c (in the ( x , t ) plane) for the equation: M ( x , t ) + N ( x , t ) x / ( t ) = 0, then, when taking the total differential of φ ( x , t ) = c : d φ ( x , t ) = 0 = φ / t dt + φ / x dx 0 = φ / t + φ / x d x / d t = M ( x , t ) + N ( x , t ) x / ( t ) M ( x , t ) = φ / t , N ( x , t ) = φ / x . So because of the continuity of φ ( x , t ) : 2 φ / t x = 2 φ / x t M / x = N / t . In the specific case above: N ( x , t ) = r ( t ), M ( x , t ) = [ p ( t ) x ( t ) – g ( t )] r ( t )

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2 Example (The Bernoulli Equations) A first order Bernoulli Equation is of the form: ( ) ( ) ( ) ( ) ( ) t x t q t x t p t x n = + & (for any integer n > 1) This equation is not linear, as one of its terms (i.e. on the right hand side) involves a power (greater than 1) of the unknown function. It can be made linear by first diving both sides by x n ( t ) and then using the substitution: u ( t ) = x (1-n) ( t ) = x ( t ) x -n ( t ): ( ) ( ) ( ) ( ) ( ) q px x x t x t q t x t p t x n n n = + = + - - 1 & & (dividing through by x n ( t )). Using substitution u ( t ) = x (1-n) ( t ) = x ( t ) x -n ( t ), note: ( ) ( ) ( ) ( ) ( ) ( ) t x t x n t u t x t u n n & & - - - = = 1 1 Hence the equation becomes: ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) t q t u t p t u q px x x t x t q t x t p t x n n n n = + = + = + - - - & & & 1 1 1 (which is linear) Consider the Bernoulli Eqn: ( ) ( ) ( ) t x t t x t x t 2 2 3 = + & .
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MA360Handout 6b - MA360 Sp2006 Kallfelz Handout 6b(week of...

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