Question

(1 pt) A. Find y in terms of x if dydx=x4y−3 and y(0)=5. y(x)= . B. For what x-interval is the solution defined? (Your answers should be numbers or plus or minus infinity. For plus infinity enter "PINF"; for minus infinity enter "MINF".) The solution is defined on the interval:

Answer 1

Answer:

$y=(\frac{4}{5}x^5+625)^{\frac{1}{4}}$

y=f(x)\ x E R (x is all real numbers)

Step-by-step explanation:

A. We have the differential equation

$\frac{dy}{dx}=x^4y^{-3}$

by using separation of variables we have

$y^3dy=x^4dx\\\\\int y^3dy=\int x^4dx\\\\\frac{1}{4}y^4=\frac{1}{5}x^5+C'\\\\y=(\frac{4}{5}x^5+C)^{\frac{1}{4}}$

C is the new constant C=4C'

$y(0)=(C)^{\frac{1}{4}}=5\\C=625$

hence y is

$y=(\frac{4}{5}x^5+625)^{\frac{1}{4}}$

B.

the interval is calculated by taking into account

$y=f(x)^{\frac{1}{4}}=f(x)^{\frac{1}{2}}f(x)^{\frac{1}{2}}$

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!!