Integrating Factor

The integrating factor is used to solve first-order linear ODEs of the form:

$$\frac{dy}{dx} + P(x)y = Q(x)$$

First, we find the integrating factor, which is given by:

$$\mu(x)=e^{\int P(x)~dx}$$

Next, we multiply both sides of the original ODE by the integrating factor:

$$\mu(x)\frac{dy}{dx}+\mu(x)P(x)y=\mu(x)Q(x)$$

Recognize that the left-hand side is the result of the product rule, so can be rewritten as:

$$\left(\mu(x)y(x) \right)'=\mu(x)Q(x)$$

Now, integrating with respect to $x$ gives:

$$\mu(x)y(x)=\int \mu(x)Q(x)~dx + C$$

Finally, we can solve for $C$. If given an initial condition, find $C$ and solve for $y(x)$.

Example

Solve the following ODE:

$$\frac{dy}{dx} + 2y = e^{-x}$$

First, we identify $P(x) = 2$ and $Q(x) = e^{-x}$. Next, we find the integrating factor:

$$\mu(x) = e^{\int 2~dx} = e^{2x}$$

Now, we multiply both sides of the original ODE by the integrating factor:

$$e^{2x}\frac{dy}{dx} + 2e^{2x} y = e^{2x}e^{-x}$$

This simplifies to:

$$\left(e^{2x}y\right)' = e^{x}$$

Now, we integrate both sides with respect to $x$:

$$e^{2x}y = \int e^{x}~dx + C$$

This gives us:

$$e^{2x}y = e^{x} + C$$

Finally, we can solve for $y(x)$:

$$y(x) = e^{-x} + Ce^{-2x}$$

If we are given an initial condition, we can find $C$ and write the final solution. For example, if we are given $y(0) = 1$, we can substitute $x=0$ and $y=1$ into the solution to find $C$:

$$1 = e^{0} + Ce^{0} \implies C = 0$$

Thus, the final solution to the ODE with the initial condition is:

$$y(x) = e^{-x}$$