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Solve for the particular solution: y' = [e^(-y)]/[t+1] where y(0) = 0?
This is differential equations, how do I get the particular solution?
2 Answers
- BrianLv 77 years agoFavorite Answer
This is a separable d.e., so rewrite as
e^y dy = dt/(t + 1) and then integrate both sides to get
e^y = ln lt + 1l + C ----->
y(t) = ln (ln lt + 1l + C).
Now apply the initial condition y(0) = ln (ln(1) + C) = ln(C) = 0,
which gives us C = 1 and thus y(t) = ln(ln lt + 1l + 1).
- NiallLv 77 years ago
This equation can be solved through separation of variables, rewrite it as:
dy/dt = e^(-y) / (t + 1)
Multiply both sides by e^(y) and multiply the differential across:
e^(y) * dy = 1 / (t + 1) * dt
Integrate both sides:
e^(y) = ln(t + 1) + C
Make y the subject, take the natural log of both sides:
y = ln[ln(t + 1) + C]
Now sub in the initial condition to solve for C:
0 = ln[ln(1) + C]
0 = ln(C)
C = 1
y = ln[ln(t + 1) + 1]
