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Unformatted text preview: te, u2 = x2 + C , • determine C using the initial condition: when x = 1, u(1) = y (1) = 1, 12 = 1 + C ,
C = 0, so, u2 = x2
• isolate u = y , y = u = ±x, Note that y > 0 at x = 1, so we pick the positive branch,
y = x
• Integrate, y = 1 x2 + c1 , determine c1 using the initial condition, y (1) = 1 12 + c1 = 1,
2
2
c1 = 1
2
1
1
y = x2 +
2
2
4. [5 marks] Find the general solution to y − 2y + y = ex
.
x • This is not in a form that the undetermined coeﬃcients method applies. Use variation
of parameters
• The solution of homogeneous part: – The auxiliary equation: λ2 − 2λ + 1 = 0, λ = 1, repeated roots.
– two linearly independent solutio...
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This test prep was uploaded on 03/31/2014 for the course MATH 201 taught by Professor Steacy during the Fall '10 term at University of Victoria.
 Fall '10
 STEACY
 Math

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