Respuesta :
Answer:
A.[tex]y(x) =\sqrt{ 2(\frac{x^5}{5} -\frac{6^5}{5})}[/tex]
B.Therefore the solution is defined on the interval:
6≤ x ≤ +∞
Step-by-step explanation:
Given differential equation is
[tex]\frac{dy}{dx} = x^4y^{-1}[/tex]
[tex]\Rightarrow \frac{dy}{dx} =\frac{ x^4}{y}[/tex]
[tex]\Rightarrow y dy = x^4dx[/tex]
Integrating both sides:
[tex]\Rightarrow\int y dy = \int x^4dx[/tex]
[tex]\Rightarrow \frac{y^2}{2} =\frac{x^5}{5} +C[/tex]........(1)
Given y(0)= 6
it means y=0 when x=6
Putting x=6 and y=0 in the equation (1)
[tex]\therefore \frac{0^2}{2} =\frac{6^5}{5} +C[/tex]
[tex]\Rightarrow C=-\frac{6^5}{5}[/tex]
Equation (1) become:
[tex]\Rightarrow \frac{y^2}{2} =(\frac{x^5}{5} -\frac{6^5}{5})[/tex]
[tex]\Rightarrow y} =\sqrt{ 2(\frac{x^5}{5} -\frac{6^5}{5})}[/tex]
Therefore [tex]y(x) =\sqrt{ 2(\frac{x^5}{5} -\frac{6^5}{5})}[/tex]
(B)
The quantity under root must greater than or equal to zero.
Therefore,
[tex]2(\frac{x^5}{5} -\frac{6^5}{5})} \geq0[/tex]
[tex]\Rightarrow \frac{x^5}{5} -\frac{6^5}{5}} \geq0[/tex]
[tex]\Rightarrow \frac{x^5}{5} \geq\frac{6^5}{5}}[/tex]
[tex]\Rightarrow x^5\geq {6^5[/tex]
[tex]\Rightarrow \sqrt[5]{x^5} \geq \sqrt[5]{6^5}[/tex]
[tex]\Rightarrow x\geq {6[/tex]
When x≥0 then the value of [tex]2(\frac{x^5}{5} -\frac{6^5}{5})}[/tex] is real other wise the value of [tex]2(\frac{x^5}{5} -\frac{6^5}{5})}[/tex] is imaginary.
Therefore the solution is defined on the interval:
6≤ x ≤ +∞
