Consider the following initial value problem Consider the following initial valu

Question

Consider the following initial value problem

Consider the following initial value problem: dx/dt = 3t + 3tx x(0) = 1 (a) Solve the initial value problem using the separable DE method. (b) Solve the same IVP (initial value problem) using the linear DE method. (c) Consider now the IVP dx/dt = 3t + 3tx x(0) = -1 which is the same differential equation x' = 3t + 3tx but with a different initial value x(0) = -1. What is the solution of this initial value problem? (d) Does the solution for IVP in (1) reach the value x = 2 for some t? (e) Does the solution for IVP in (2) reach the value x = 2 for some t?

Explanation / Answer

According to Chegg policy, only four subquestions will be answered. Please post the remaining in another question

(a) dx/dt = 3t + 3tx

=> dx/dt = 3t (1 + x)

=> dx 1/(1+x) = 3t dt

Integrating both sides,

ln (1+x) = 3t2/2 + c

=> 1 + x = e(3t2/2 + c)

=> 1 + x = e3t^2/2 ec

=> 1 + x = ke3t^2/2 where k = ec

=> x = ke3t^2/2 - 1

Given x(0) = 1

=> 1 = ke0 - 1

=> 1 = k - 1

=> k = 2

The solution is x = 2e3t^2/2 - 1

(c) Using the common form above,

x = ke3t^2/2 - 1

given x(0) = -1

-1 = ke0 - 1

=> k = 0

The solution is x = -1

(d) x = 2e3t^2/2 - 1

When x = 2

2 = 2e3t^2/2 - 1

=> e3t^2/2 = 3/2

=> 3t2/2 = ln(3/2)

=> t2 = 2/3 ln(3/2)

thus the solution does reach the value x = 2 for some t

(e) Since the solution is a constant, the value x = 2 is never reached.

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