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# How to Solve a Differential Equation Using Integrating Factors

### First-Order Nonhomogeneous Differential Equations

In physics and engineering problems, differential equations of the form

y' + f(x)y = g(x)

often arise. These first-order nonhomogeneous equations may look difficult at first glance, but they can be quickly solved by multiplying both sides of the equation by an *integrating factor*.

The integrating factor converts the left-hand side into a derivative, so the whole problem can be solved by integration.

### The Method

First, multiply both sides of the diff. eq. by the function e^{F(x)}, where F(x) = ∫f(x) dx, or equivalently F'(x) = f(x).

e

^{F(x)}y' + e

^{F(x)}f(x)y = e

^{F(x)}g(x)

Notice that the left-hand side is now the derivative of e

^{F(x)}y by the product rule for differentiation. This means that we can rewrite the equation as

[e

^{F(x)}y]' = [∫e

^{F(x)}g(x) dx]'

And now it can be solved by integration and simple algebra:

e

^{F(x)}y = ∫e

^{F(x)}g(x) dx

y = [∫e

^{F(x)}g(x) dx]/e

^{F(x)}

### Example 1

Solve y' + 4x^{3}y = e

^{x-x4}. In this case, the integrating factor is e

^{x4}since the integral of 4x

^{3}is x

^{4}. So we have

y' + 4x

^{3}y = e

^{x-x4}

e

^{x4}y' + e

^{x4}4x

^{3}y = e

^{x}

[e

^{x4}y]' = [e

^{x}]'

e

^{x4}y = e

^{x}+ c

y = e

^{x-x4}+ ce

^{-x4}

### Example 2

Solve y' - (1/x)y = ln(x). First, since the anti-derivative of -1/x is -ln(x), we determine that the integrating factor is e^{-ln(x)}= 1/x.

y'/x - y/x

^{2}= ln(x)/x

y/x = (1/2)ln(x)

^{2}+ c

y = (x/2)ln(x)

^{2}+ cx

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