
Let
, so that
:

Now the ODE is separable, and we have

Integrating both sides gives

For the integral on the left, rewrite the integrand as

Then

and so


Given that
, we find

so that the particular solution to this IVP is

The equation of the line would be y = (1/2)x.
To find the equation of a line that is reflected through the line y = x, we just switch the x and y. Then, solve for y.
y = 2x
x = 2y
0.5x = y
If the equation is r = 3 +4cos(θ) then because b/a>1 the curve is a limacon with an inner loop.
Given limacon with equation r=3+4cos(θ) and we have to answer how the quotient of a and b relate to the existence of an inner loop.
Equation is like a relationship between two or more variables expressed in equal to form and it is solved to find the value of variables.
formula of polar graph is similar to r= a+ b cos (θ).
Case 1. If a<b or b/a>1
then the curve is a limacon with inner loop.
Case 2. If a>b or b/a<1
Then the limacon does not have an inner loop.
Here given that
(θ)
It is observed that , a<b or b/a>1
Therefore the curve is limacon with an inner loop.
Hence because b/a>1 the curve is a limacon with an inner loop.
Learn more about limacon at brainly.com/question/14322218
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Her correct change would be $2.75 cents, to get this answer, just subtract 12.42 from 15.17
Hope this helped you!! ( :