Answer: t = - 5 ∈ [tex]I_{1}[/tex] = ( -∞ , -4 )
Step-by-step explanation:
The standard form of O.D.E is written as :
[tex]y^{1}[/tex] + [tex]p(t) = g(t)[/tex]
Equation given :
[tex](16-t^{2} )y^{1}[/tex] + [tex]9ty[/tex] = [tex]5t^{2}[/tex] , [tex]y(-5) = 1[/tex]
The first thing to do is to write the O.D.E in standard form , that is we will divide through by [tex]16 - t^{2}[/tex] , so we have
[tex]y^{1} + \frac{9ty}{16-t^{2}}=\frac{5t^{2}}{16-t^{2}}[/tex]
With this , we can see that [tex]p(t)[/tex] and [tex]g(t)[/tex] are both continuous in the same domain. Therefore , the intervals are :
[tex]I_{1}[/tex] = ( -∞ , -4 )
[tex]I_{2}[/tex] = ( - 4 , 4 )
[tex]I_{3}[/tex] = ( 4 , -∞ )
recall that y(−5) = 1 , then t = -5
This means that :
t = - 5 ∈ [tex]I_{1}[/tex] = ( -∞ , -4 )
