Question

Solve the initial-value problem.
tdu/dt= t2 + 3u, t > 0, u(4) = 48

Answer 1

To solve the initial-value problem tdu/dt = t^2 + 3u for t > 0 with the condition u(4) = 48, one must apply mathematical techniques such as separation of variables or an integrating factor to find the solution that fits the given condition.

Step-by-step explanation:

The initial-value problem given is tdu/dt = t2 + 3u, with initial condition u(4) = 48, and defined for t > 0. This is a first-order linear ordinary differential equation which can be solved using separation of variables or an integrating factor. Since acceleration formulas and velocity are used as references, it is clear that this question pertains to mathematical methods used in physics, specifically in dynamics and kinematics.

Nonetheless, to solve the differential equation provided in the question, one would proceed by first separating variables or using an integrating factor to get u as a function of t, then applying the initial condition to find the particular solution that satisfies u(4) = 48.

Answer 2

The initial-value problem is solved by finding an integrating factor and using separation of variables which leads to the solution u(t) = t^2 - t + t^3.

Step-by-step explanation:

The differential equation t du/dt = t2 + 3u needs to be solved with the initial condition that u(4) = 48. This is a first-order linear differential equation and can be solved by separating variables or using an integrating factor. For separation of variables, divide both sides by t and rearrange the terms to separate the u terms from the t terms:

du/dt - (3/t)u = t

Next, find an integrating factor μ(t) such that μ(t) = exp(-3 ∫ dt/t) = exp(-3 ln|t|) = t-3. Multiplying through by this factor yields:

t-3 du/dt - 3t-4 u = t-2

The left side is now the derivative of (u/t3) with respect to t, so we can integrate both sides with respect to t:

∫ d(u/t3) = ∫ t-2 dt

This yields:

u/t3 = -(1/t) + C

Now, plug in the initial condition u(4) = 48 to solve for C:

48/43 = -(1/4) + C

C = 48/64 + 1/4 = 3/4 + 1/4 = 1

Thus, the solution to the differential equation is:

u(t) = t3(—1/t + 1)

Or more simply:

u(t) = t2 - t + t3