# chapter2 - Chapter 2 Solutions to 2.1 1 x2 and ex are both...

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Chapter 2 Solutions to §2.1 1. x 2 and e x are both continuous functions for all x R and in particular on any interval I containing x 0 = 0. y(x) = 0 is clearly a solution to the given DE and also satisfies the initial conditions. Thus, by the existence–uniqueness theorem, y(x) = 0 is the only solution to the initial value problem. 3. If y 1 (x) = cos 2x, and y 2 (x) = sin 2x then y 1 ´´ + 4y 1 = (cos 2x)´´ + 4cos 2x = –4cos 2x + 4cos 2x = 0 and y 2 ´´ + 4y 2 = (sin 2x)´´ + 4sin 2x = –4sin 2x + 4sin 2x = 0. Hence, cos 2x and sin 2x are solutions to the DE. Further, W[cos 2x, sin 2x](x) = (cos 2x)(2cos 2x) – (sin 2x)(–2sin 2x) = 2cos 2 (x) + 2sin 2 (x) = 2 0, so {cos 2x, sin 2x} is a linearly independent set of solutions to the DE on any interval and it follows from Theorem 2.1.4 that the general solution is given by y(x) = c 1 cos 2x + c 2 sin 2x. 5. x 2 y´´ + 3xy´ – 8y = x 2 (x α )´´ + 3x(x α )´ – 8x α = x 2 [ α ( α – 1)x α – 2 ] + 3x( α x α – 1 ) – 8x α = α ( α – 1)x α + 3 α x α – 8x α = 0. Now x α > 0 so α(α – 1) + 3 α – 8 = 0 ⇒α 2 + 2 α – 8 = 0 α = 2 or α = –4. Hence y 1 (x) = x 2 , y 2 (x) = x –4 are solutions to the given DE. Further, W[x 2 , x –4 ] = (x 2 )(–4e –5 ) – (x –4 )(2x) = –6x –3 0, so that {x 2 , x –4 } is a linearly independent set of solutions to the given DE on (0, ). Consequently, from Theorem 2.1.4, the general solution is given by y(x) = c 1 x 2 + c 2 x –4 . 7. a) y´´ + y´ – 2y = 0 (e rx )´´ + (e rx )´ – 2e rx = 0 r 2 e rx + re rx – 2e rx = 0 r 2 + r – 2 = 0 r = –2 or r = 1. Thus y 1 (x) = e –2x , y 2 (x) = e x are solutions to y´´ + y´ – 2y = 0. Further, these solutions are linearly independent on any interval since W[e –2x , e x ] = (e –2x )(e x ) – (e x )(–2e –2x ) = 3e –x 0. It follows that y c (x) = c 1 e –2x + c 2 e x is the complementary function for the given DE. 57

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58 Ch. 2 Second-Order Linear Differential Equations b) y´´ + y´ – 2y = 4x 2 (a 0 + a 1 x + a 2 x 2 )´´ + (a 0 + a 1 x + a 2 x 2 )´ – 2(a 0 + a 1 x + a 2 x 2 ) = 4x 2 (a 1 + 2a 2 x)´ + (a 1 + 2a 2 x) – 2(a 0 + a 1 x + a 2 x 2 ) = 4x 2 2a 2 + a 1 + 2a 2 x – 2a 0 – 2a 1 x – 2a 2 x 2 = 4x 2 (2a 2 + a 1 – 2a 0 ) + (2a 2 – 2a 1 )x – 2a 2 x 2 = 4x 2 . Equating like terms we obtain the system 2a 2 + a 1 – 2a 0 = 0, 2a 2 – 2a 1 = 0, and –2a 2 = 4 which has the solution a 0 = –3, a 1 = –2, and a 2 = –2. Thus, y p (x) = –(3 + 2x + 2x 2 ). c) From Theorem 2.1.5, the general solution is y(x) = c 1 e –2x + c 2 e x + 2e 3x – (3 + 2x + 2x 2 ). Solutions to §2.2 1. Given that y 1 (x) = x 2 . Let y 2 (x) = x 2 u so y 2 ´ = 2xu + x 2 u´, and y 2 ´´ = x 2 u´´ + 4xu´ + 2u. Substituting these results into the original DE and simplifying we obtain xu´´ + u´ = 0 u´´ = 1 x u(x) = c 1 ln x + c 2 . Letting c 1 = 1 and c 2 = 0 we obtain a second linearly independent solution, y 2 (x) = x 2 ln x. 3. Given that y 1 (x) = e x . Let y 2 (x) = ue x so that y 2 ´ = e x u + e x u´, and y 2 ´´ = e x u + 2e x u´ + e x u´´. Substituting these results into the original equation and simplifying we obtain xu´´ + u´ = 0 u´´ = 1 x = c 1 x u(x) = c 1 ln x + c 2 . Letting c 1 = 1 and c 2 = 0, we obtain a second linearly independent solution, y 2 (x) = e x ln x.
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