Determine the longest interval in which the given initial value problem is certain to have a unique twice-differentiable solution. Do not attempt to find the solution. (Enter your answer using interval notation.) ty'' + 7y = t, y(1) = 1, y'(1) = 7

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.

since proposition states that a number is greater than 25, if is true then the number must actually be greater than 25.

this implies that all possible options, except 24, can be accepted

ddsf

step-by-step explanation:

dfs