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differential equations - B U Department of Mathematics Math...

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B U Department of Mathematics Math 202 Differential Equations Date: June 2, 2004 Full Name : Time: 15:00-17:30 Math 202 Number : Student ID : Spring 2004 Final Exam Solution Key IMPORTANT 1. Write your name, surname on top of each page. 2. The exam consists of 7 questions some of which have more than one part. 3. Read the questions carefully and write your answers neatly under the corresponding questions. 4. Show all your work. Correct answers without sufficient explanation might not get full credit. 5. Calculators are not allowed. Q1 Q2 Q3 Q4 Q5 Q6 Q7 total 20 pts 20 pts 20 pts 20 pts 20 pts 25 pts 25 pts 150 pts 1.) [20] Solve the following differential equation by finding an integrating factor μ = μ ( x ): ( x + 2) sin y dx + x cos y dy = 0 . Solution: Realize that ∂M ∂y = ( x + 2) cos y = cos y = ∂N ∂x and the differential eqution is not exact. Then p ( x ) = ( x + 2) cos y - cos y x cos y = x + 1 x = 1 + 1 x and μ ( x ) = e p ( x ) dx = e (1+ 1 x ) dx = e x +ln x = xe x . So multiplying by this integrating factor, μ ( x ) = xe x , the following differential equation: ( xe x ( x + 2) sin y ) dx + ( x 2 e x cos y ) dy = 0 is exact. Let Φ = ( x 2 e x cos y ) dy = x 2 e x sin y + ψ ( x ). Then Φ ∂x = ψ ( x ) + 2 xe x sin y + x 2 e x sin y = xe x ( x + 2) sin y = x 2 e x sin y + 2 xe x sin y. So ψ ( x ) = 0 and ψ ( x ) = c a constant. The family of solutions is : x 2 e x sin y + c = 0.
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2.) [20] Find the general solution of the differential equation: y - y = 3 t + cos t. Solution: The characteristic equation is r 3 - r = 0. Then we get three distinct roots. r = 0 , +1 , - 1. The homogeneous solution is y h ( t ) = c 1 + c 2 e t + c 3 e - t . To find the particular solution we use the the method of undetermined coefficients. Let y p ( t ) = At + Bt 2 + C cos t + D sin t . Then y p ( t ) = A + 2 Bt - C sin t + D cos t, y p ( t ) = 2 B - C cos t - D sin t, y p ( t ) = C sin t - D cos t.
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