Webwork 9.1/Problem 3
Solve the differential equation
du
= e2u+5t .
dt
for the initial condition u(0) = 7. (In your case the numbers may have differed.)
We separate the variables: e−2u du = e5t dt and integrate (already bringing both constants to one side):
1
1
− e−2u = e5t + C.
2
5
Solving for u yields
1 5t
1
e +C
u = − ln −2 ·
2
5
Do not worry about the −2 in the argument for the logarithm – we will have C a big negative number such
that the argument of the logarithm is positive!
Now we use the initial condition to get the value for C. Setting t = 0 yields:
1 5·0
1
e +C
7 = u(0) = − ln −2 ·
2
5
and therefore
1
1
1
1
C = − e−2·7 − e5·0 = − e−14 − .
2
5
2
5
Thus the solution we want is:
1 5t 1 −14 1
1
e − e
−
.
u = − ln −2 ·
2
5
2
5
We see from this solution that for t > 51 ln 2e514 + 1 ∼ 4.1 · 10−7 the argument of the logarithm becomes
negative. This means that for larger t values the differential equation has no solution.
This is not surprising if we plot the solution — we see that at this t-value the solution curve goes to ∞.
If the differential equation came from a model in physics, this essentially means that the model does not
work for larger t-values for the given initial configuration u(0) = 7. (For example at this point the machine
will have been broken or the particle has moved away or hit a wall.)
1