Exercise 1.6
1. Solve the following differential equations.
(a)
(b)
(c)
(d)
2. Solve the following initial value problems.
(a)
(b)
(c)
(d)
3. Given RL circuit with .
Find i(t).
1.
(a) This is a linear diffenretial equation. Now write this in the standard form.
Now the integrating factor is given by
Multiply to the standard form. Then
Now the left-hand side is the derivative of the times the dependent variable . Then
Integrate both sides by .
Therefore, the integrating factor is
(b) This equation is a linear differential equation. So, the integrating factor is given by
Multiply this to the standard form, we have
Now the left-hand side is the derivative of times .
Integrate both sides with respect to , we have
Therefore, the general solution is
(c) This equation is a linear differntial equation. So, write in the standard form. Then
The integrating factor is
Multiply to the standard form. Then we have
Now the left-hand side is the derivative of times .
Integrate both sides with respect to .
Therefore, the general solution is
(d) This equation is a linear differential equation. Now write in the standard form. Then
Integrating factor is
Multiply this to the standard form.
Then the left-hand side is the derivative of times .
Integrate both sides with respect to .
Therefore, the general solution is
2.
(a) This equation is a linear differential equation. The integrating factor is given by
Multiply this to the standard form.
The left-hand side is the derivative of times .
Integrate both sides with respect to .
Thus, the general solution is
Now by the initial condition , we have
and
(b) This equation is a linear differential equation. So, write in the standard form.
The integrating factor is
Multiply this to the standard form.
The left-hand side is the derivative of times .
Integrate both sides by . Then
Therefore, the general solution is
BY the initial value , we have
and
(c) This equation is a linear differential equation. So, the integrating factor is given by
Multiply this to the standard form.
Then the left-hand side is the derivative of times .
Integrate both sides by , we have
Therefore, the general solution is
Now by the initial value , we have
and
.Thus, for
,
. Note that the solution of this differential equation must be continuous. Then
and
.Thus,
. From this, we have
(d) This equation is a linear differential equation. Then the integrating factor is
Now multiply to the standard form. Then
The left-hand side is the derivative of the product of and .
Integrate both sides by .
Therefore, the general solution is
Now using initial condition , we have
and
3. The differential equation for the current running through RL-circuit is
Then we have the standard form.
It is a linear differential equation. So, we find the integrating factor.
Multiply this to the standard form.
The left-hand side is the derivative of the product of and .
Integrate both sides with respect to .
Then the general solution is
Now we use the initial condition . Then we have
and
. Therefore,
Next we consider the case . Since , the differential equation for the current running through RL-circuit is
and