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By Elliott Ward Cheney

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The other two solutions are defined on t < 1 and t < 2/3, respectively. Notice also that each becomes unbounded as t approaches 1/(2a2 ) from the left. 3). On the other hand, a nonlinear equation may have an implicit solution that cannot be solved analytically for the dependent variable to find an explicit form of the solution. of dy/dt = y1/5 is y = 45 t + C or y = 0, solve the problem by hand. Is the solution unique? 1 4 t both satisfy 4. Show that y = 0 and y = 16 √ the IVP 1/t dy/dt = y, y(0) = 0.

Theorem 3. Let f (t, y) be continuous. A function φ(t), defined on an interval I, is a solution of y = f (t, y), y(t0 ) = y0 if and only if it is a continuous t solution of y = y0 + t0 f (s, y)ds. Prove this theorem. Hints: First assume that φ(t) is a solution of y = dy/dt = f (t, y), y(t0 ) = y0 . This means φ = f (t, φ). Now integrate both t t sides of the equation t0 φ (s)ds = t0 f (s, φ(s))ds and simplify using the Fundamental Theorem of Calculus. For the converse, differentiate φ(t) = t y0 + t0 f (s, φ(s))ds and simplify.

16. 17. 18. 19. 20. 21. d2 x/dt2 + 4x = 0 d2 x/dt2 − 5 dx/dt = 0 d2 x/dt2 + 4 dx/dt + 13x = 0 x − 6x + 7x = 0 ( = d/dt; x = x(t)) x + 16x = sin t x + 4x + 13x = e−t (a) Use a computer algebra system to solve 2 dy/dx = e−x ; (b) Graph the solution to the 2 IVP dy/dx = e−x , y(0) = a for a = −2, −1, 0, 1, 2. (c) Graph the slope field of the differential equation together with the solutions in (b). 2 (a) For any given initial conditions, in a brief paragraph explain why you think that the solutions of the systems are similar or different.

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