Finding a particular solution of a differential equation

I already solved the homogeneous portion, and I need help solving the particular solution and of course combining the two to get the entire solution to the differential equation. Not too difficult - see attachment. Please use equation editor if possible. Thank you. --- Given that: dMS/dt = m(MN - MS) - pMS¬ MN¬ (0) = MS (0) = 0 And using: M¬N + MS = [Po/(r + p)](ert - e-pt) Show that MS(t) = BY SOLVING THE DIFFERENTIAL EQUATION FOR dM¬s¬/dt with a homogeneous and particular solution. Using the MN + MS equation to solve for MN¬ in terms of MS gives t... click for more

Solving a Particular solution to an Ordinary Differential equation

See attached problem. PLEASE NOTE!!! I have noted in the problem statement that I have solved the homogeneous portion of the differential equation, and I need assistance in solving for the particular solution and finally the whole solution. I have gotten 2 responses from other OTA's that are as follows: "The point is that you have NO NEED to solve the differenmtial equation - because you are already given the answer. All you have to do is to use the answer for M_S to calculate M_N (from the second equation), then differentiate M_S with respect to t, substitute all you got in the differential equation and show that it is satisfied. Do not forget to check the M_N (0) = M_ (0) = 0 ... click for more

Local/uniform Lipschitz constants

Determine if the following functions satisfy local or uniform Lipschitz condition. 1). te^y My work: I found d/dy (te^y) = te^y, and this is not bounded above for any value of y, so this made me conclude that it has locally Lipschitz condition since the Lipschitz constant here changes as the reagion changes? Am I right? I used the equation | f(t,y_1) - f(t, y_2) | = d/dy(f(t,y)) + | y_1 - y_2| If my work is incorrect, provide the correct answer and approach. 2). y t^2/ (1 + y^2) I used the same approach here and d/dy (f ) = t^2 - 2y + y^2/ ( 1 + y^2)^2, which is clearly could be bounded above by a constant but this constant changes as the reagion changes so it is local ... click for more

Initial value problem for systems of DE

Let Q(t) =< (less than or equal) C + integral from t_0 to t ( K(s) Q(s) ) ds, Where Q(t) is a nonegative function , C > 0 and K(s) >= 0. a).Show that: Q(t) =< Ce^( integral from t_0 to t ( K(s)ds) ), t >= t_0 b). What conclusion can be made if C = 0? ( Note that proof in a may fail is C = 0 ). I want a detailed proof, please justify every claim you make. Thanks.

Systems of equations as transformations

Let X be a normed space, I closed interval ( or half-open on the right) and a = inf I, b = sup I. Let h : I -> [0,infinity) be a continuous function such that integral ( from a to b ) h(t)dt < positive infinity where integral from a to b represents the improper integral when I is not closed. Let epsilon > 0 and X_epsilon,h be the set of all continuous functions f : I -> X such that the number ||f||_e,h defined by ||f||_epsilon,h = (for t in I) sup( e^-(epsilon*integral(from a to t ) h(s)ds)) ||f(t)|| is finite. Prove that the when the interval I is compact, the set X defined above of normed spaces coincides with C(I,X), and that the sup norm and the norm ||.||_epsi... click for more