In Problems 1 through 4, use Theorem 5 to discuss the existence and uniqueness of a solution to the differential equation that satisfies the initial conditions y(1)=Yo,y'(1)=Y1, whereYo and Y1are real constants.

t(t-3)y''+2ty'-y=t2

Short Answer

Expert verified

The differential equation has a unique solution in 0<t<3.

Step by step solution

01

Find the value of p(t), q(t),g(t).

The given differential equation ist(t-3)y''+2ty'-y=t2

It can be written as y''+2tt(t-3)y'-yt(t-3)=t2t(t-3).

So,p(t)=2tt(t-3),q(t)=-1t(t-3),g(t)=t2t(t-3)

02

Check the result

Here p(t), q(t),g(t) are continuous functions in the interval 0<t<3 and 2<t< and the pointt0=1 in the continuity interval 0<t<3.

Therefore, the differential equation has a unique solution in 0<t<3.

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