Question

[IMAGE INCLUDED] 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?

Answer 1

C)
`j=6; k=1`

Explanation:

`y=-ln(j-ke^(2x))\\\\-ln(5)=-ln(j-ke^(2(0)))\\\\-ln(5)=-ln(j-k)\\\\ln(5)=ln(j-k)\\\\5=j-k`

`y=-ln(j-ke^(2x))\\\\y=ln((1)/(j-ke^(2x)))\\\\(dy)/(dx)=-(2ke^(2x))/(ke^(2x)-j)\\ \\2e^(2x+y)=-(2ke^(2x))/(ke^(2x)-j)\\\\2e^(2(0)+(-ln(5)))=-(2ke^(2(0)))/(ke^(2(0))-j)\\\\2e^(-ln(5))=-(2k)/(k-j)\\ \\(2)/(5)=-(2k)/(k-j)\\\\-10k=2k-2j\\\\-12k=-2j\\\\6k=j`

`5=j-k\\\\5=6k-k\\\\5=5k\\\\1=k`

`6k=j\\\\6(1)=j\\\\6=j`

Therefore, j=6 and k=1