Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Differential Equations test one solutions Tuesday, April 25, 2006 x Z 1. Given the function y = 4+ sin s2 ds 2 2 Z (a) Calculate y(2) = 4+ (b) Calculate sin s2 ds = 4+0 = 4 2 2. Is the function y = x+ dy = sin x2 dx 1 dy 2 a solution of +y = 3+x2 ? Yes because: x dx dy d 1 1 + y2 = x+ + x+ dx dx x x 1 1 = 1 − 2 + x2 + 2 + 2 x x 2 = 3 + x2 3. Determine whether each of the following equations is linear or non-linear. (a) dy = t − y2 dt non-linear (b) t dy = et y−cos t2 dt (c) y linear dy = 4t − y dt non-linear 4. Which of the following equations is exact? (a) 3x2 − 4y dx+(2y − 4x) dy = 0 (b) dy 2 +x y = x or (x2 y−x)dx+(1)dy = 0 dx 2 3x − 4y ? y = (2y − 4x)x x2 y − x ? y = (1)x x2 6= 0 not exact −4 = −4 exact page two 5. Given the initial value problem t dy 3 4 + y= and y(3) = 5 dt t − 2 t+1 determine the largest t-interval over which a unique solution is guaranteed to exist. Rewrite the differential equation as dy 3 4 + y= dt t(t − 2) t(t + 1) The coefficients are not continuous at t = 0 and t = 2. The solution to the initial value problem exists until a discontinuity is reached, so that the solution will exist on the interval t > 2. 6. Given the initial value problem dy √ = y 3 + x 3 y and y(3) = 0 dx can one be certain that a unique solution exists? ∂ 3 y + xy 1/3 = y=0 ∂y x 3y + √ 3 3 y2 2 ! = undefined y=0 Therefore a unique solution is not certain to exist. 7. Use Euler’s method (i.e. the tangent-line method) with step-size h = 1/2 to fill in the following table for the initial value problem dy = 2x − y 2 and y(1) = 2 dx x y dy dx 1 2 2(1) − 22 = −2 (−2)(1.5 − 1) = −1 1.5 2 + (−1) = 1 2(1.5) − (1)2 = 2 (2)(0.5) = 1 ∆y ≈ dy ∆x dx page three 8. Solve each of the following differential equations or initial value problems: (a) dy = 2xey dx separable =⇒ Z Z Z dy Z −y = 2x dx =⇒ e dy = 2x dx =⇒ −e−y = x2 +C ey y = − ln(−x2 + C 0 ) where C 0 = −C (b) dy dt 2t dy + 2y = 5e3t y(0) = 3 Z linear =⇒ find integrating factor: µ = exp 2 dt = e2t . Multiply equation: Z d 2t 5t 2t e + 2e y = 5e e =⇒ e y = 5e =⇒ e y = 5e5t dt = e5t + C dt dt 2t 2t 3t Plug in the initial conditions to find that C = 2, and solve for y: y = e3t + 2e−2t (c) 3x2 − 4y dx+(2y − 4x) dy = 0 This equation is exact (see problem 4a). A solution is of the form F (x, y) = C, where F (x, y) = Z 2 3 3x − 4y dx = x − 4xy + g(y) and F (x, y) = Therefore the solution is x3 − 4xy + y 2 = C. Z 2y − 4x dy = y 2 − 4xy + h(x) page four 9. The rate of population growth of a small country is proportional to the country’s population. At time t = 0 years the country’s population is equal to 5 million. (a) From time t = 0 years to time t = 20 years the population grows at an instantaneous rate equal to 4% of the population. Set up the differential equation for the country’s population P (t) and solve it. dP = .04P =⇒ P = 5e.04t when 0 ≤ t ≤ 20 dt (b) Starting at time t = 20 years the population grows at an instantaneous rate equal to only 2% of the population. Set up the differential equation for the country’s population during this interval and solve it. dP = .02P =⇒ P = Ce.02t = 5e.04(20) e.02(t−20) = 5e.02(t+20) when t ≥ 20 dt