#### Method 1 - Solving the DE

The formula is: `Ri+L(di)/(dt)=V`

After substituting: `50i+(di)/(dt)=5`

We re-arrange to obtain:

`(di)/(dt)+50i=5`

This is a first order linear differential equation.

We'll need to apply the formula for solving a first-order DE (see Linear DEs of Order 1), which for these variables will be:

`ie^(intPdt)=int(Qe^(intPdt))dt`

We have `P=50` and `Q=5`.

We find the integrating factor:

`"I.F."=e^(int50dt)=e^(50t)`

So after substituting into the formula, we have:

`(i)(e^(50t))=int(5)e^(50t)dt` `=5/50e^(50t)+K` `=1/10e^(50t)+K`

When `t=0`, `i=0`, so `K=-1/10=-0.1`.

This gives us: `i=0.1(1-e^(-50t))`

The transient current is: `i=0.1(1-e^(-50t))\ "A"`.

The steady state current is: `i=0.1\ "A"`.

#### Method 2: Using the Formula

NOTE: We can use this formula here only because the voltage is **constant**. This formula will not work with a variable voltage source.

We have the following general formula:

`i=V/R(1-e^(-(R"/"L)t))`

So in this case:

`i=5/50(1-e^(-50t))` `=0.1(1-e^(-50t))`

Graph of the current at time `t`, given by `i=0.1(1-e^(-50t))`.

In this example, the time constant, TC, is

`tau=L/R=1/50=0.02`

So we see that the current has reached steady state by `t = 0.02 \times 5 = 0.1\ "s".`

#### Method 3: Using Scientific Notebook's Solve ODE

If you have Scientific Notebook, proceed as follows:

This DE has an initial condition *i*(0) = 0. We set up a matrix with 1 column, 2 rows.

For the answer: **Compute** → **Solve ODE...** → **Exact**