# It turns out that the function f x x m is a solution

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It turns out that the function f ( x ) = x m is a solution to the differential equation precisely when 0 2 x ( mx m 1 ) x 2 ( m ( m 1) x m 2 ( m ( m 3)) x m for all x. This is equivalent to m = 0 or m = 3. ______________________________________________________________________ 3. (5 pts.) It is known that every solution to the differential equation y - y = 0 is of the form . Which of these functions satisfies the initial conditions y (0) = 2 and y (0) = 8 ?? The initial conditions lead to the system of equations which is equivalent to The solution to the IVP is given by .

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TEST1/MAP2302 Page 3 of 3 ______________________________________________________________________ ______________________________________________________________________ 4. (10 points) The following differential equation may be solved by either performing a substitution to reduce it to a separable equation or by performing a different substitution to reduce it to a homogeneous equation. Display the substitution to use and perform the reduction, but do not attempt to solve the separable or homogeneous equation you obtain . (5 x 2 y 1) dx (2 x y 1) dy 0 is equivalent to The substitution is x = X + 1 and y = Y - 3. The reduction results in the homogeneous DE (5 X 2 Y ) dX (2 X Y ) dY 0 _________________________________________________________________ Bonkers 10 Point Bonus: (a) The Fundamental Theorem of Calculus provides a neat formal solution involving a definite integral with respect to the variable t to the following dinky IVP: y ( x ) e x 2 and y (0) 1. What is that solution? (b) Unfortunately the function g ( x ) e x 2 cannot be integrated in elementary terms. Use the answer to (a), the Maclaurin series for e x , and term-by-term integration, to obtain a power series solution to the IVP. Write your answer using sigma notation. [Say where your work is! You don’t have room here!] (a) y ( x ) 1 x 0 e t 2 dt for all x . (b) y ( x ) 1 x 0 e t 2 dt 1 x 0 k 0 ( t 2 ) k k ! dt 1 k 0 x 0 ( t 2 ) k k ! dt 1 k 0 x 0 t 2 k k ! dt 1 k 0 x 2 k 1 (2 k 1) k ! for all x .
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