(X+2)(x-4)(x+7)(x-1)
1 answer:
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Answer:
For this case we can use the theorem of Existence and uniqueness that says:
Let p(t) , q(t) and g(t) be continuous on [a,b] then the differential equation given by:
has unique solution defined for all t in [a,b]
If we apply this to our equation we have that p(t) =0 and and
We see that is not defined at t =0, so the largest interval containing 1 on which p,q and g are defined and continuous is given by
And by the theorem explained before we ensure the existence and uniqueness on this interval of a solution (unique) who satisfy the conditions required.
Step-by-step explanation:
For this case we have the following differential equation given:
With the conditions y(1)= 1 and y'(1) = 7
The frist step on this case is divide both sides of the differential equation by t and we got:
For this case we can use the theorem of Existence and uniqueness that says:
Let p(t) , q(t) and g(t) be continuous on [a,b] then the differential equation given by:
has unique solution defined for all t in [a,b]
If we apply this to our equation we have that p(t) =0 and and
We see that is not defined at t =0, so the largest interval containing 1 on which p,q and g are defined and continuous is given by
And by the theorem explained before we ensure the existence and uniqueness on this interval of a solution (unique) who satisfy the conditions required.