Answer:
[tex]y=(\frac{4}{5}x^5+625)^{\frac{1}{4}}[/tex]
y=f(x)\ x E R (x is all real numbers)
Step-by-step explanation:
A. We have the differential equation
[tex]\frac{dy}{dx}=x^4y^{-3}[/tex]
by using separation of variables we have
[tex]y^3dy=x^4dx\\\\\int y^3dy=\int x^4dx\\\\\frac{1}{4}y^4=\frac{1}{5}x^5+C'\\\\y=(\frac{4}{5}x^5+C)^{\frac{1}{4}}[/tex]
C is the new constant C=4C'
[tex]y(0)=(C)^{\frac{1}{4}}=5\\C=625[/tex]
hence y is
[tex]y=(\frac{4}{5}x^5+625)^{\frac{1}{4}}[/tex]
B.
the interval is calculated by taking into account
[tex]y=f(x)^{\frac{1}{4}}=f(x)^{\frac{1}{2}}f(x)^{\frac{1}{2}}[/tex]
in both factors f(x) can take negative values because i*i=-1 a real number. Hence we have the interval
y=f(x)\ x E R (x is all real numbers)
hope this helps!!
