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midterm1

# midterm1 - 18.034 Midterm#1 Name Part I TF Questions Answer...

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Unformatted text preview: 18.034 Midterm #1 Name: Part I: TF Questions. Answer for each of the following statements if it is true or fale. Simply say T (if you believe it is true) or F (if you suspect it is false); you don’t need to justify your answers. Each question counts 5pts. 1. The solution of the DE xy + 2 y = 4 x 2 for x > with the initial condition y (1) = 2 has the minimum value 2 . 1 + 3 x 2 2. The solution of the initial value problem y = 3 y 2 − 6 y , y (0) = 1 is defined for any x ∈ ( −∞ , ∞ ) . 3. The DE x 2 + x y dx − x 2 + y y dy = 0 is exact 2 2 4. The solution of the DE u u − x = 0 with the initial conditions u (1) = 2 ,u (1) = 1 is defined for all x ∈ ( −∞ , ∞ ) . 5. Every solution of the Bessel equation of order zero x 2 u + xu + x 2 u = 0 vanishes infinitely often on (0 , ∞ ) . 6. If two functions f and g are linearly independent on the interval (0 , 1) , then they are linearly independent on (0 , 2) . 1 18.034 Midterm #2 Name: Part II: Proofs. Problem 1. We consider the initial value problem (1) y = f ( x, y ) , y (0) = . If f and ∂f/∂y are continuous in a rectangle | x | ≤ a , | y | ≤ b , then there is some interval | x | ≤ h ≤ a in which there exists a unique solution y = φ ( x ) of (1). In this problem we indicate how to prove this. In order to prove the existence part, it is necessary to transform the initial value problem (1) into a more convenient form. If we suppose for now that there is a function y = φ ( x ) which satisfies the initial value problem, then f ( x, φ ( x )) is a continuous function of a single variable x , and hence we integrate the differential equation against x to obtain the integral equation x (2) φ ( x ) = f ( s, φ ( s )) ds....
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midterm1 - 18.034 Midterm#1 Name Part I TF Questions Answer...

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