Exam 1 solution

# Exam 1 solution - PRELIM 1 MATH 293 SPRING 2006 STUDENT'S...

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PRELIM 1 , MATH 293 , SPRING 2006 STUDENT’S NAME: TA’S NAME: PROBLEM 1: PROBLEM 2: PROBLEM 3: PROBLEM 4: PROBLEM 5: TOTAL: 1

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2 PRELIM 1, MATH 293, SPRING 2006 Problem 0.1. Find the general solution of the following differential equation y 4 dy dx = ( y 5 + 3) sin x. Proof This is separable because the right-hand side of dy dx = y 5 + 3 y 4 sin x is a product of a function of x and a function of y . Therefore, y 4 y 5 + 3 dy = sin x dx. To get the integral on the left, let u = y 5 + 3, so then du = 5 y 4 dy , and y 4 y 5 + 3 dy = 1 5 u du. Evaluating the integrals, 1 5 ln | u | = 1 5 u du = - cos x + C. Taking exp of both sides, | u | 1 / 5 = e - cos x + C . Raising both sides to the 5th power, u = ± ( e - cos x + C ) 5 = ke - 5 cos x . Substituting back for u , y 5 + 3 = ke - 5 cos x , so y = ( ke - 5 cos x - 3) 1 / 5 .
PRELIM 1, MATH 293, SPRING 2006 3 Problem 0.2. Show that the substitution v = ln y transforms the differential equation dy/dx + P ( x ) y = Q ( x )( y ln y ) into the linear equation dv/dx + P ( x ) = Q ( x ) v ( x ) . Then, use this observation to solve the differential equation x dy dx - 5 x 3 y + 3 y ln y = 0 . Proof If v = ln y then y = e v , so dy dx = e v dv dx by the chain rule. The differential equation transforms into xe v dv dx - 5 x 3 e v + 3 e v v = 0 .

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