Solve x\'(t) = 2 - tx^2 - t + 2x^2 with x(0) = 0. What, exactly, does Theorem 1,
ID: 2881704 • Letter: S
Question
Solve x'(t) = 2 - tx^2 - t + 2x^2 with x(0) = 0. What, exactly, does Theorem 1, Existence and Uniqueness of Solutions, on page 22 in Section 1.3 of the text, tell you about the solution which you found in question #3? As always, pretend that you have graduated, and are working somewhere, and one of your colleagues has asked you to explain this to them. Subject your answer to the will-this-person-be-back-in-fifteen-minutes-for-further-clarification test, and the is-there-any-way-that-my-explanation-could-be-misunderstood test.Explanation / Answer
from the given question,
x'(t) = 2- tx2 - t + 2x2
dx/dt= (2-t) + x2( 2-t)
dx/dt= (2-t) (1+x2)
dx/ (1+x2) = (2-t) dt
arc tanx = (2t - t2/2 ) +c
replacing initial conditions,
x(0)=0, x=0, t=0
arc tan0 = (0-0) +c
c=0
arc tanx = (2t - t2/2 )
x= tan (2t - t2/2 )