Dy dx 2 x y x c this little first order linear

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dy dx 2 x y x . (c) This little first order linear equation has µ = x 2 as an integrating factor. Using the standard recipe, a one-parameter family of solutions is given by . x 2 y x 4 4 K Note: The Student Solutions’s Manual has some nonsense in its printed solution of this varmint.
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TEST2/MAP2302 Page 4 of 4 ______________________________________________________________________ 7. (10 pts.) It turns out that the nonzero function f( x ) = exp( x) is a solution to the homogeneous linear O.D.E. (*) y y 0. (a) Reduction of order with this solution involves making the substitution y ve x into equation (*) and then letting w = v . Do this substitution and obtain the constant coefficient equation that w must satisfy. (b) Obtain a the general solution to the ODE that w satisfies and then stop. (c) Explain very briefly why v can be obtained from w without actually integrating. Do not attempt to actually find v. (a) If we have , y ve x then , y e x v e x v , y e x v 2 e x v e x v and . y e x v 3 e x v 3 e x v e x v Substituting y into (*) and then replacing v using w implies that w must be a solution to (**) w 3 w 3 w 0. (b) The auxiliary equation for (**) above is m 2 3 m 3 0. By using quadratic formula, it is easy to see that the roots are m 3 2 ± 3 2 i The general solution to (**), then, is w c 1 cos( 3 2 x ) e 3 2 x c 2 sin( 3 2 x ) e 3 2 x . (c) The equation v = w is a constant coefficient linear ODE with w being a UC function. [ Look at w above, Folks! ] Consequently, we can completely solve this ODE without performing any actual integrations. _________________________________________________________________ Silly 10 Point Bonus: Let f ( x ) = x and g ( x ) = sin( x ). (a) It is trivial to obtain a 4th order homogeneous linear constant coefficient ordinary differential equation with f and g as solutions. Do so.
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