# 2. Solve the initial value problem. 11(t + 1) dy dx − 9y = 18t, for t > 0

```2. Solve the initial value problem.
11(t + 1)
dy
− 9y = 18t,
dx
for t &gt; 0 with y(0) = 6.
Solution: Dividing both sides by 11(t + 1), we rewrite the differential equation
as follows:
18 t
9
y0 −
y=
(1)
11(t + 1)
11 t + 1
so we can identify the coefficient
P (t) = −
9
.
11(t + 1)
We then find the integrating factor,
I(t) = e
R
P (t) dt
9
= e− 11
R
1
t+1
dt
9
= e− 11 ln(t+1) = (t + 1)−9/11 .
Multiply both sides of equation (1) by I(t) to get
(t + 1)−9/11 y 0 −
18
9
(t + 1)−9/11−1 y =
(t + 1)−9/11 (t + 1)−1 t.
11
11
Notice the left hand side is the derivative of the product of I(t) and y. So,
simplifying, we have
0
18
t
(t + 1)−9/11 y =
.
11 (t + 1)20/11
Now integrate both sides. On the left, this is simply (t+1)−9/11 y. You integrate
the right-hand side as I showed you in class. So, we have,
11
18 11
2/11
−9/11
−9/11
(t + 1)
+ (t + 1)
+C
(t + 1)
y=
11 2
9
= 9(t + 1)2/11 + 2(t + 1)−9/11 + C.
Now, here is where I made a mistake in class. When you solve for y, be
careful to multiply everything on the right-hand side by (t + 1)9/11 , even C, as
follows:
y = 9(t + 1)9/11 (t + 1)2/11 + 2(t + 1)9/11 (t + 1)−9/11 + (t + 1)9/11 C
= 9(t + 1) + 2 + (t + 1)9/11 C.
(2)
Finally, you must find C so that y(0) = 6. That is,
6 = y(0) = 9(0 + 1) + 2 + (0 + 1)9/11 C = 11 + C.
So, C = −5, (as we correctly computed in class). Plug this value for C into
equation (2) to get the final correct solution:
y = 9(t + 1) + 2 − 5(t + 1)9/11 .
1
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