I'm having trouble with this Calculus question, can anyone help?

Question

The solution to dy/dx = 2e^(2x+y) with initial condition (0, –ln 5) can be expressed as y = –ln(j – ke^(2x)) for real numbers j and k. What are the values of j and k?

A. j = 5, k = 1

B. j = 5, k = 2

C. j = 6, k = 1

D. j = 7, k = 2

Answer 1

Let's reference dy/dx as equation A and y = -ln(...) as equation B.

Use the fact that (0,-ln5) is a point on the graph of equation B to get one equation involving j and k. This part is fairly straightforward.

-ln(5) = -ln(j-ke²⁽⁰⁾) => 5 = j - k.

Finding another equation involving j and k might work as follows.

Find the derivative from equation B and set it equal to the right side of equation A.

Realize that you can write the right side of equation A as:

2e^{(2x+ (-ln(j-ke^2x)} = 2e^2xe^{(-ln(j-ke^2x)}

After setting that equal to the derivative found from equation B, using some properties of logarithms, along with some algebraic simplification....

Give that a try and post a comment if you still need some help.