Values of J and K

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?

j = 5; k = 1

j = 5; k = 2

j = 6; k = 1

j = 7; k = 2

Answer 1

dy/dx = 2e^2x . e^y ===> dy / e^y = 2e^2x dx.

Now use the change of variable as:

e^y = u and e^y dy = du or dy / e^y = du/u².

Therefore:

du/u² = 2e^2x dx

where by integrating both sides and rearrangement we obtain:

y = -ln(-c - e^2x ).

Now replace (0, –ln 5) in above to find j and k.