Practice Test #1 MAT 242
(1) Identify (a) the order of the DE, (b) the dependent and independent variables, (c) whether the DE is linear or non-linear:
(a)
(b)
(2) Draw the slope field for the DE, and classify the equilibrium points as stable or unstable:
(3) Tell whether or not the function is a solution of the equation .
(4) Show that the relation is an implicit solution to the differential equation (tip: use implicit differentiation).
(5) Does theorem 1 imply the existence of a unique solution to the following DE's?
(a)
(b)
(6) Use an integrating factor to solve the differential equation:
(a)
(b)
(7) Solve the differential equation using separation:
(a)
(b)
(8) Verify that the equation is exact, and solve:
(2xy2 + 2y) + (2x2y + 2x)y' = 0.
(9) At time t = 0, a tank contains Q0 lb. of salt dissolved in 120 gallons of water. Assume that water containing 1/3 of salt/gal. is entering the tank at a rate of 2 gal/min. and is exiting at the same rate. Set up the initial value problem that models this process, with Q(t) equal to the amount of salt in the tank at any given time t, and find the limiting amount Ql that is present after a long time.
Solutions:
(1) (a) order = 3, (b) dependent = y, independent = t, (c) linear
(b) (a) order = 2, (b) dependent = y, independent = x, (d) non-linear (yy')
(2) p = 10 is unstable, p = 5 is stable
(3)
looks good – solution.
(4) diff both sides, keeping in mind that you have to 'chain on' when diff'ing terms involving y:
there you go.
(5)
(a) yes – the right-hand side and are both continuous at the initial value coordinates (1,0)
(b) no - the right-hand side is not continuous at the initial value coordinates (1,0) ('cuzz you can't divide by 0).
(6)(a) First, get standard form by dividing through by x:
then you need to identify P(x) , and find the integrating factor:
Multiply both sides by the integrating factor, etc:
(b) is in standard form, so indentify P and find the integrating factor:
multiply through by the integrating factor, etc.:
(7) (a) multiply by dx, and divide by y:
integrate both sides, let it rip:
(b) multiply by dx, and multiply through by y:
integrate both sides, and simplify:
(8) (2xy2 + 2y) + (2x2y + 2x)y' = 0, so M(x,y) = 2xy2 + 2y, and N(x,y) =2x2y + 2x, and
Nx = 4xy + 2 = My, and that means that the equation is exact. To find the implicit solution, we take the partial differential equations
fx (x,y) = 2xy2 + 2y, and fy (x,y) = 2x2y = 2x, and integrate both sides of each to get :
f(x,y) = x2y2 + 2xy + h(y) , and f(x,y) = x2y2 + 2xy + h(x), and unioning the terms we have:
f(x,y) = x2y2 + 2xy = c as our implicit solution
(9) so change in salt level = rate in - rate out, or
Q' = (1/3)2 - 2(Q/120), and Q(0) = Q0 (Q/120 is the percentage of water in the tank). Put this in the form:
Q' + (1/60)Q = 2/3, and the integrating factor is e∫(1/60)dt =e (1/60)t, and so we have the solution:
Q = (∫ e (1/60)tdt + C)/ e (1/60)t = ( 60e (1/60)t + C)/ e (1/60)t , so finally
Q = 60 + C e -(1/60)t , and now use the IC Q(0) = Q0 to get
Q0 = 60 + C e -(1/60)0, and so C = Q0 - 60,
Q = 60 + (Q0 - 60) e -(1/60)t, and if you take the limit as t approaches infinity , it appears this goes to Q = 60 lbs.