# The Method of Integrating Factors Examples 3

Recall from The Method of Integrating Factors page that we can solve first order linear differential equations of the form $\frac{dy}{dt} + p(t) y = g(t)$ by multiplying both sides of the equation by the integrating factor $\mu (t) = e^{\int p(t) \: dt}$ so that:

(1)

We will now look at some examples of applying this method of integrating factors.

## Example 1

**Solve the initial value problem. $t \frac{dy}{dt} +2y = t^2 – t + 1$ for $t > 0$ and with the initial condition $y(1) = \frac{1}{2}$.**

We first need our differential equation in the appropriate form. We divide each side by $t$ to get:

(2)

We now see that $p(t) = \frac{2}{t}$. Therefore our integrating factor is:

(3)

We now multiply both sides of our differential equation by our integrating factor $\mu (t)$ and get:

(4)

Plugging in our initial condition of $y(1) = \frac{1}{2}$ and we have that:

(5)

Therefore the solution to this initial value problem is:

(6)

## Example 2

**Solve the initial value problem. $t^3 \frac{dy}{dt} + 4t^2 y = e^{-t}$ for $t and with the initial condition $y(-1) = 0$.**

Once again, we will divide each term in our differential equation, this time by $t^3$, to get it in the appropriate form:

(7)

We see that $p(t) = \frac{4}{t}$ and so our integrating factor is:

(8)

We now multiply both sides of our differential equation by $t^4$ to get:

(9)

The integral on the righthand side can be evaluated with integration by parts. Let $u = t$ and $dv = e^{-t} \: dt$. Then $du = dt$ and $v = -e^{-t}$. Therefore we have that:

(10)

Therefore we have that:

(11)

Plugging in the initial condition of $y(-1) = 0$ and we see that $C = 0$. Therefore the solution to this initial value problem is:

(12)

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