7. Solve the following differential equations :3e

Question

x
tan y dx + (1 – e
x
) sec2 y dy = 0 given that , 4
y
�
 when
x = 1.

Answer 1

wow. That copy/paste doesn't work too well, eh? If I can figure it out, we have

3e^x tany dx + (1+e^x)sec^2y dy = 0
(1+e^x)sec^2y dy = -3e^x tany dx

sec^2y/tany dy = -3e^x/(1+e^x) dx
1/(siny cosy) dy = -3e^x/(1+e^x) dx
2csc(2y) dy = -3e^x/(1+e^x) dx
ln(tany) = -3ln(1+e^x) + c
tany = c/(1+e^x)^3
y = arccot(c(1+e^x)^3)

Now just plug in your point to find c.

Of course, if I misinterpreted your text, you may have to modify the solution shown.