Question

The IVP

sin(t)d2xdt2+cos(t)dxdt+sin(t)x=tan(t)

x(1)=17

dxdt∣∣1=8

has a unique solution defined on the interval

Answer 1

Given: ах sin t +cos tー+xsint=tan t at dt Initial conditions: x(1)= 17 and x'(t)=8 Solution: Rewrite the differenti al equati on, we have dx cost dr xsint tan t dtsint dt sinf sin t sint sin t dbc -cott-+ x = drt sin t ax +cot t + x = sec t We know that, the second order DE is of the form Comparing this with (1), we have p (t) = cot t ,q (t1-1 and r (t) sec t Here, p(t) is defined for all values of t except So that, p(t) is continuous at possible intervals are q(t) is defined for all values of t except So that, g(t) is continuous at the possible intervals are "( 2.2八2.2八2.2八2.2

(-즉위 ππ Therefore, the pos sible intervals containing x = 1 are (0, π) and So that, pl).a(t) and r(t) continuous at 0 So that, p(t)-g(t) and r(t) continuous at | 0,二 Thus, the unique solution defined on the interval is 0