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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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