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

Answer:

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.

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

Step-by-step explanation:

a. Solving as a separable equation.

Divide both sides by y:

$\frac{1}{y} 2t\frac{dy}{dt}=6y\frac{1}{y} \\\\\frac{2t}{y}\frac{dy}{dt} =6$

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

$\frac{dy}{y} =\frac{3}{t}dt$

Integrate both sides:

$\int\ \frac{dy}{y} =\int\ \frac{3}{t} dt$

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:

$F(t,y)=\frac{6y}{2t} \\\\and\\\\\frac{\partial F}{\partial y} =\frac{6}{2t} }$

The domain of F(t,y) is:

$t\in R \hspace{12}t\neq0$

The domain of  $\frac{\partial F}{\partial y}$  is:

$t\in R \hspace{12}t\neq0$

So, Existence and Uniqueness theorem tells us that for each $t\in R:\hspace{12}t\neq0$ 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:

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