Homework-2-s11-solution

Homework-2-s11-solution - 1 AMS 361 Applied Calculus IV...

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Unformatted text preview: 1 AMS 361: Applied Calculus IV Homework 2 Spring 2011 Problem 2-1 : Find the general solutions (implicit if necessary, explicit if convenient) of the differential equations in the following problem: ( + 2) 2 ′ = ( + 2) 2 Solution: If ≠ − 2 , the equation is transformed: ( + 2) 2 = ( + 2) 2 Take one step of anti-derivative action: − 1 + 2 ¡ = ( − 1 + 2 ) ⇒ 1 + 2 ¡ = ( 1 + 2 ) 1 + 2 = 1 + 2 + = + 2 + 1 + 2 One more step gives the general solution: = + 2 + 2 + 1 − 2 = (1 − 2 ) − 4 + 2 + 1 If = − 2 , it satisfies the original equation. It is one solution, too. The final solution is combined by two parts above. Problem 2-2 : Find the explicit particular solution of the initial value problem ¢ ′ = £ 2 ¤ = /2 Solution: The solution is given below: 2 = ⇒ = =...
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Homework-2-s11-solution - 1 AMS 361 Applied Calculus IV...

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