Question

Consider the initial value problem 2ty′=6y, y(−2)=−8.

a. Find the value of the constant C and the exponent r so that y=C(t^r) is the solution of this initial value problem.
y= ?

b. Determine the largest interval of the form a
c. What is the actual interval of existence for the solution (from part a)?

Answer 1

a.

`C=1\\r=3\\y=t^3`

b.

`(-\infty,0)\cup(0,\infty)\\\\\\-\infty<t<0\hspace{3}\cup\hspace{3}0<t<\infty`

c.

`(-\infty,\infty)\\\\-\infty<t<\infty`

Explanation:

a. Solving as a separable equation.

Divide both sides by y:

`(1)/(y) 2t(dy)/(dt)=6y(1)/(y) \\\\(2t)/(y)(dy)/(dt) =6`

Divide both sides by 2t and multiply both sides by dt:

`(dy)/(y) =(3)/(t)dt`

Integrate both sides:

`\int\ (dy)/(y) =\int\ (3)/(t) dt`

Evaluate the integrals:

`log(y)=log(t)+C`

Where C is an arbitrary constant.

Solving for y:

`y(t)=e^(C) t^3=Ct^3\\Because\hspace{3}e^(C)\hspace{3}is\hspace{3}in\hspace{3}fact\hspace{3}another\hspace{3}constant`

Evaluating the initial condition:

`y(-2)=-8=C(-2)^3\\\\-8=C(-8)\\\\C=1`

So:

`y(t)=t^3`

b.

Let:

`F(t,y)=(6y)/(2t) \\\\and\\\\(\partial F)/(\partial y) =(6)/(2t) }`

The domain of F(t,y) is:

`t\in R \hspace{12}t\\eq0`

The domain of
`(\partial F)/(\partial y)` is:

`t\in R \hspace{12}t\\eq0`

So, Existence and Uniqueness theorem tells us that for each
`t\in R:\hspace{12}t\\eq0` there exists a unique solution defined in an open interval around
`t`

`(-\infty,0)\cup(0,\infty)\\\\\\-\infty<t<0\hspace{3}\cup\hspace{3}0<t<\infty`

c. The domain of y(t) is:

`y\in R`

Hence, the actual interval of existence for the solution y(t) is:

`(-\infty,\infty)\\\\-\infty<t<\infty`