midterm2-A-solution-public

# Yp a1 2xe2x yp a4x 4e2x yp yp 2yp

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Unformatted text preview: olution: yp = axe−2x (we need x multiplied to e−2x because e−2x is a solution to the homogeneous part). – – – – ￿ ￿￿ yp = a(1 − 2x)e−2x , yp = a(4x − 4)e2x , ￿￿ ￿ yp + yp − 2yp = a(4x − 4)e−2x + a(1 − 2x)e−2x − 2axe−2x = −3ae−2x = 3e−2x − 3 a = 3, a = − 1 So, yp = −xe−2x • The general solution is y = yp + c1 y1 + c2 y2 = −xe−2x + c1 ex + c2 e−2x • Alternatively, use variation of parameters – y = u1 y 1 + u2 y 2 , ￿ ￿ – The Wronskian W = y1 y2 − y1 y2 = ex (−2e−2x ) − ex e−2x = −3e−x u￿1 u￿2 y2 f ( x) e − 2x 3 e − 2x =− =− = e − 3x −x W − 3e y1 f ( x) e x 3 e − 2x = = = −1 W − 3 e −x 1 – Integrate: u1 = ˆ 1 e−3x dx = − e−3x + c1 3 u2 = ˆ −1dx = −x + c2 – So, the general solution is 1 y1 = − e−3x ex + c1 ex − xe−2x + c2 e−2x 3 1 = −xe−2x + (c2 − )e−2x + c1 ex 3 = −xe−2x + c1 ex + c3 e−2x where c3 = c2 − 1/3. 3. [5 marks] Find the solution of the initial value problem y ￿ y ￿￿ = 4x, y (1) = 5, y ￿ (1) = 2 • Let u = y ￿ , y ￿￿ = u￿ , then, uu￿ = 4x, • separable for...
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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.

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