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AMath350.W14.A3.Sol

# 2 r dy 2y e x an integrating factor is e 2dx e2x

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Unformatted text preview: n integrating factor is e 2dx = e2x . The equation dx d 2x can be rewritten as e y = ex , and so we ﬁnd that dx a) For y=e 2x (ex + C ) = e x + Ce 2x . We can see that y = e x is an exceptional solution, and all other solutions approach it as x ! 1, and are repelled by it as x ! 1. AMATH 350 Page 3 Assignment #3 Solutions Next, observe that y 0 = 0 if and only if y = 1 e x (this is our horizontal 2 isocline). The solutions must look something like this: 3 2 1 -5 -4 -3 -2 -1 C>0 ← C=0 0 -1 C<0 -2 Figure 1: dy b) For x = 3x 2 dx form: 1 2 3 4 5 ↖ All solutions with C<0 eventually cross the x-axis, increasing until they cross the horizontal isocline, and then decrease alongside the exceptional solution. -3 y we need to start by putting the equation in standard dy 1 + y = 3x. dx x R 1 An integrating factor is µ(x) = e x dx = eln x = x. Incorporating this, we obtain dy x + y = 3x2 dx which must be the same equation as dy (xy ) = 3x2 . dx Therefore xy = x3 + C and so C . x Now, y = x2 is an exceptional solution. All other solutions approach this as x ! 1 and diverge from it as x ! 0 every solution has a vertical asymptote at the origin (except for the exceptional solution). The horizontal isocline is y = 3x2 . Finally, notice that C > 0 makes y > x2 if x > 0, but y < x2 if x < 0. Putting our observations together, we arrive at the following sketch: y = x2 + AMATH 350 Page 4 Assignment #3 Solutions 3 C<0 C>0 2 1 ← C=0 -5 -4 -3 -2 -1 0 C>0 1 -1 2 3 4 5 C<0 -2 Figure 2: -3 3. (Interest Rate Problem) Let’s let y be the balance of the account, and measure t in years. For continuously compounded interest alone, the balance would evolve according to the dy Mathusian model: = 0.05y . To account for continuous depositing at the dt rate of \$365 per annum, we simply add a term: dy = 0.05y + 365. dt The initial condition is y (0) = 1000. The DE is linear, and an integrating factor is µ=e R 1 dt 20 t/20 =e . Factoring this into the equation, we obtain e t/20 dy 1 e 20 dt t/20 y = 365e t/20 which must be equivalent to d e dt t/20 = 365e t/20 (and we can quickly check that it is, so we have calculated our integrating factor correctly). Integrating, we ﬁnd e so t/20 y= t/20 7300e y...
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