# First order linear differential equations

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Say we have

${a_1}(x)y' + {a_0}(x)y + b(x) = 0$

We will rewrite it to

$y' + \frac{{{a_0}(x)}}{{{a_1}(x)}}y = - \frac{{b(x)}}{{{a_1}(x)}}$

or

$y' + p(x)y = q(x)$

This is usually the form it is expressed in.  Ok. How to solve this? To figure that out we will work a bit backwards. We will somewhat do a trick similar to completing the square.  Let us look at a special case of the product rule

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

This is kind of almost what we have on the LHS of our equation. Would it not be nifty if we could make it have that form? Then we could collapse it to a single derivative, and then we could take the antiderivative of both sides to find our solution. What is missing is some factor in front of y´.  So let’s multiply through with some factor I. We get

$Iy' + I \cdot p(x)y = I \cdot q(x)$

So, to be able to collapse it to what we want to have we must have that

$f(x) = I\quad f'(x) = I \cdot p(x)$

Combining the two equations above we get

$I' = I \cdot p(x)$

or

$\frac{{dI}}{I} = p(x)dx$

We can solve this to get

$\ln |I| = \int {p(x)dx = P(x) + {C_1}}$

or

$I = {e^{P(x) + {C_1}}} = {C_2}{e^{P(x)}}$

where, as usual, we let the constant absorb the sign.

Back to our equation, multiplying through be the above we get

${C_2}{e^{P(x)}}y' + {C_2}{e^{P(x)}} \cdot p(x)y = {C_2}{e^{P(x)}} \cdot q(x)$

As we can see we can immediately divide away the constant, so we could instead have used

$I = {e^{P(x)}} = {e^{\int {p(x)dx} }}$

to get

${e^{P(x)}}y' + {e^{P(x)}} \cdot p(x)y = {e^{P(x)}} \cdot q(x)$

This could now be collapsed to

$({e^{P(x)}}y)' = {e^{P(x)}} \cdot q(x)$

I.e. the first term without the derivative.  Integrating (taking the antiderivative) of both sides gives us

${e^{P(x)}}y = \int {{e^{P(x)}} \cdot q(x)dx}$

or

$y = \frac{{\int {{e^{P(x)}} \cdot q(x)dx} }}{{{e^{P(x)}}}}$

or, written out with all our integrals in full glory

$y = \frac{{\int {{e^{\int {p(x)dx} }} \cdot q(x)dx} }}{{{e^{\int {p(x)dx} }}}}$

Quite a formidable formula that you thankfully normally are not supposed to remember. Instead you remember that you have to multiply through by

$I = {e^{\int {p(x)dx} }}$

that we will call the integrating factor.

Up a level : Differential Equations
Previous page : Linear differential equations
Next page : First order linear differential equations - a couple of examplesLast modified: Mar 18, 2019 @ 18:18