Determine (without solving the problem) an interval in which the solution of the given initial value problem is certain to exist. (t−3)y′+(ln⁡t)y=2t, y(1)=2

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Determine (without solving the problem) an interval in which the solution of the given initial value problem is certain to exist. (t−3)y′+(ln⁡t)y=2t, y(1)=2

Jordan Mitchell · Accepted Answer

Step 1Divide equation with (t−3),y′+ln⁡tt−3 y=2tt−3According to Theorem 2.4.1,p(t)=ln⁡tt−3 and g(t)=2tt−3Functions p(t), g(t) are continuos on intervals (0, 3) and (3, ∞) because domains are (0, ∞){3} for p and R{{3}} for g.Since t0=1, solution exists on (0, 3)

Paul Mitchell · Accepted Answer

First we put the ODE into standard form:y′+ln⁡(t)(t−3)y=2t(t−3)Then we see thatp(t)=ln⁡(t)(t−3),whileg(t)=2t(t−3)Now,p(t)=ln⁡(t)(t−3)is continuous on the  intervalsIp1=(0, 3)and Ip2=(3, +∞)whileg(t)=2t(t−3)is continuous on the intervalsIg1=(−∞, 3)andIg2=(3, +∞)Thus we see that both p(t) and g(t) are continuous on Ip1 and Ip2=Ig2. However, the initial condition is at t=1, y(t)=2; t=1 is in the intervalIp1=(0, 3)thus we are guranteed a solution on the interval Ip1=(0, 3)

nick1337 · Accepted Answer

Step 1Rewriting the above equation in the standard form, we havey′+ln⁡tt−3y=2tt−3So p(t)=ln⁡tt−3 and g(t)=2tt−3.g is continuous for all t≠3. p is continuous for all t≠0, 3.Therefore p and g are both continuous on the interval (−∞, 0)∪(0, 3)∪(3, ∞)The interval (0, 3) contains the initial point t=1. Therefore Theorem 3.1 guarantees that the problem has  a unique solution on the interval 0&amp;lt;t&amp;lt;3We now turn our attention to nonlinear differential equations and modify Theorem 3.1 by a more general theorem.

Determine (without solving the problem) an interval in which the

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