Answer:
I accidently clicked answer and it won't let me quit
Step-by-step explanation:
I'm not even in high school yet D: sorry
Answer:
B the range, the x- and y-intercept
Step-by-step explanation:
the domain stays the same : all values of x are possible out of the interval (-infinity, +infinity).
but the range changes, as for the original function y could only have positive values - even for negative x.
the new function has a first term (with b) that can get very small for negative x, and then a subtraction of 2 makes the result negative.
the y-intercept (x=0) of the original function is simply y=1, as b⁰=1.
the y-intercept of the new function is definitely different, because the first term 3×(b¹) is larger than 3, because b is larger than 1. and a subtraction of 2 leads to a result larger than 1, which is different to 1.
the original function has no x-intercept (y=0), as this would happen only for x = -infinity. and that is not a valid value.
the new function has an x-intercept, because the y-values (range) go from negative to positive numbers. any continuous function like this must therefore have an x-intercept (again, y = the function result = 0)




Answer:
y = x*sqrt(Cx - 1)
Step-by-step explanation:
Given:
dy / dx = (x^2 + 5y^2) / 2xy
Find:
Solve the given ODE by using appropriate substitution.
Solution:
- Rewrite the given ODE:
dy/dx = 0.5(x/y) + 2.5(y/x)
- use substitution y = x*v(x)
dy/dx = v + x*dv/dx
- Combine the two equations:
v + x*dv/dx = 0.5*(1/v) + 2.5*v
x*dv/dx = 0.5*(1/v) + 1.5*v
x*dv/dx = (v^2 + 1) / 2v
-Separate variables:
(2v.dv / (v^2 + 1) = dx / x
- Integrate both sides:
Ln (v^2 + 1) = Ln(x) + C
v^2 + 1 = Cx
v = sqrt(Cx - 1)
- Back substitution:
(y/x) = sqrt(Cx - 1)
y = x*sqrt(Cx - 1)