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A linear differential equation is a differential equation (either ordinary or partial) where each function and derivative of y (or any dependent variable) has an exponent of either one or zero.

For linear equations in the form

$\frac{dy}{dx} + f(x)y = g(x)$,

the solution can be found with the formula

$y = \frac{1}{\mu(x)} \int \mu(x) g(x) dx, \mu(x) = e^{\int f(x)dx}$

For example:

$\frac{dy}{dx} = 2x^2 + x^2y$
$\frac{dy}{dx} - x^2y= 2x^2$
$\frac{1}{\mu} \int (\mu(x)) (2x^2) dx, \mu(x) = e^{-\frac{1}{3} x^3}$
$2e^{\frac{1}{3} x^3} \int e^{-\frac{1}{3} x^3} x^2 dx$

By using integration by substitution, we get:

$y = 2e^{\frac{1}{3} x^3} ( -e^{-\frac{1}{3} x^3} + C ) = Ce^{\frac{1}{3} x^3} - 2$

## Proof of formula

Suppose we have a linear differential equation in the form

$\frac{dy}{dx} + f(x)y = g(x)$

Let's multiply this by an integrating factor $\mu (x) \neq 0$

$\mu(x)\frac{dy}{dx} + \mu(x)f(x)y = \mu(x)g(x)$

Let's assume that the left hand side is now integratable using the reverse product rule, and then find the conditions for $\mu(x)$ that make this possible.

The product rule says for two differentiable functions, $u,v$

$(u\cdot v)' = uv' + u'v$

In our case, $u=\mu(x)$, and $v=y$, and this means we require

$\mu'(x) = \mu(x)f(x)$

Solving this separable differential equation is simple and yields

$\mu(x) = \exp\left(\int f(x)dx \right)$

Going back to our earlier equation, we can now proceed and reverse the product rule

$\dfrac{d}{dx}\left(\mu(x)y\right) = \mu(x)g(x)$

Integrating and dividing by $\mu(x)$ gives us the formula above.

$y = \frac{1}{\mu(x)} \int \mu(x) g(x) dx, \mu(x) = e^{\int f(x)dx}$

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