Question

A. Find y in terms of x if

Answer 1

Answer:

A.$y(x) =\sqrt{ 2(\frac{x^5}{5} -\frac{6^5}{5})}$

B.Therefore the solution is defined on the interval:

6≤ x ≤ +∞

Step-by-step explanation:

Given differential equation is

$\frac{dy}{dx} = x^4y^{-1}$

$\Rightarrow \frac{dy}{dx} =\frac{ x^4}{y}$

$\Rightarrow y dy = x^4dx$

Integrating both sides:

$\Rightarrow\int y dy = \int x^4dx$

$\Rightarrow \frac{y^2}{2} =\frac{x^5}{5} +C$........(1)

Given y(0)= 6

it means y=0 when x=6

Putting x=6 and y=0 in the equation (1)

$\therefore \frac{0^2}{2} =\frac{6^5}{5} +C$

$\Rightarrow C=-\frac{6^5}{5}$

Equation (1) become:

$\Rightarrow \frac{y^2}{2} =(\frac{x^5}{5} -\frac{6^5}{5})$

$\Rightarrow y} =\sqrt{ 2(\frac{x^5}{5} -\frac{6^5}{5})}$

Therefore $y(x) =\sqrt{ 2(\frac{x^5}{5} -\frac{6^5}{5})}$

(B)

The quantity under root must greater than or equal to zero.

Therefore,

$2(\frac{x^5}{5} -\frac{6^5}{5})} \geq0$

$\Rightarrow \frac{x^5}{5} -\frac{6^5}{5}} \geq0$

$\Rightarrow \frac{x^5}{5} \geq\frac{6^5}{5}}$

$\Rightarrow x^5\geq {6^5$

$\Rightarrow \sqrt[5]{x^5} \geq \sqrt[5]{6^5}$

$\Rightarrow x\geq {6$

When x≥0 then the value of $2(\frac{x^5}{5} -\frac{6^5}{5})}$ is real other wise the value of $2(\frac{x^5}{5} -\frac{6^5}{5})}$ is imaginary.

Therefore the solution is defined on the interval:

6≤ x ≤ +∞