Question

Determine (without solving the problem) an interval in which the solution of the given initial value problem is certain to exist. (Enter your answer using interval notation.) (16 − t2)y' + 9ty = 5t2, y(−5) = 1

Answer 1

Answer: t = - 5 ∈ $I_{1}$ = ( -∞ , -4 )

Step-by-step explanation:

The standard form of O.D.E is written as :

$y^{1}$ + $p(t) = g(t)$

Equation given :

$(16-t^{2} )y^{1}$ + $9ty$ = $5t^{2}$ ,       $y(-5) = 1$

The first thing to do is to write the O.D.E in standard form , that is we will divide through by $16 - t^{2}$ , so we have

$y^{1} + \frac{9ty}{16-t^{2}}=\frac{5t^{2}}{16-t^{2}}$

With this , we can see that $p(t)$ and $g(t)$ are both continuous in the same domain. Therefore , the intervals are :

$I_{1}$ = ( -∞ , -4 )

$I_{2}$ = ( - 4 , 4 )

$I_{3}$ = ( 4 , -∞ )

recall that y(−5) = 1 , then t = -5

This means that :

t = - 5 ∈ $I_{1}$ = ( -∞ , -4 )