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

Answer:

C) $j=6; k=1$

Step-by-step explanation:

$y=-ln(j-ke^{2x})\\\\-ln(5)=-ln(j-ke^{2(0)})\\\\-ln(5)=-ln(j-k)\\\\ln(5)=ln(j-k)\\\\5=j-k$

$y=-ln(j-ke^{2x})\\\\y=ln(\frac{1}{j-ke^{2x}})\\\\\frac{dy}{dx}=-\frac{2ke^{2x}}{ke^{2x}-j}\\ \\2e^{2x+y}=-\frac{2ke^{2x}}{ke^{2x}-j}\\\\2e^{2(0)+(-ln(5))}=-\frac{2ke^{2(0)}}{ke^{2(0)}-j}\\\\2e^{-ln(5)}=-\frac{2k}{k-j}\\ \\\frac{2}{5}=-\frac{2k}{k-j}\\\\-10k=2k-2j\\\\-12k=-2j\\\\6k=j$

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

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

Therefore, j=6 and k=1