M427K-F10-HW6 - y 1 ( t ) lead to the general solution of...

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M427K , 55330, Oct 4, 2010 Homework 6 , due Oct 11 3.6.7 Find the general solution of the given di±erential equation. y ′′ + 4 y + 4 = t 2 e 2 t , t > 0 . 3.6.17 Verify that the given functions y 1 and y 2 satisfy the corresponding homogeneous equa- tion; then ²nd a particular solution of the given nonhomogeneous equation. x 2 y ′′ - 3 xy + 4 y = x 2 ln( x ) , x > 0; y 1 ( x ) = x 2 , y 2 ( x ) = x 2 ln( x ) . 3.6.28 The method of reduction of order (Section 3.4) can also be used for the nonhomoge- neous equation y ′′ + p ( t ) y + q ( t ) y = g ( t ) , ( i ) provided one solution y 1 of the corresponding homogeneous equation is known. Let y = v ( t ) y 1 ( t ) and show that y satis²es equation ( i ) if v is a solution of y 1 ( t ) v ′′ + b 2 y 1 ( t ) + p ( t ) y 1 ( t ) B v = g ( t ) . ( ii ) Equation ( ii ) is a ²rst order linear equation for v . Solving this equation, integrating the result, and then multiplying by
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Unformatted text preview: y 1 ( t ) lead to the general solution of equation ( i ). 3.6.30 Use the method outlined in Problem 28 to solve the given dierential equation. t 2 y + 7 ty + 5 y = t, t > 0; y 1 ( t ) = t 1 . 5.3.2 Determine y (0), y (0), and y (4) (0) if y is a solution of the given initial value problem. y + sin( x ) y + cos( x ) y = 0; y (0) = 0 , y (0) = 1 . 5.3.3 + Determine the rst 5 terms in the Taylor expansion about x = 1 for the solution y of the given initial value problem. x 2 y + (1 + x ) y + 3 ln( x ) y = 0; y (1) = 2 , y (1) = 0 ....
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