A harder example of solving a linear first order equation

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(This page basically follows my train of thought, so it is not all too well structured, but that is on purpose – I want you to get a feeling of the thoughts involved.)

Solving differential equations can be quite hard, and even a small change to an equation can make it all but impossible to solve. We previously found that

$y' - xy = x$

has the solution

$y = C{e^{\frac{{{x^2}}}{2}}} - 1$

Let us now instead look at

$y' - xy = {x^2}$

This is a linear first-order equation that is not separable, so we may try using an integration factor. We get

$I = {e^{\int {( - x)dx} }} = {e^{ - {x^2}/2}}$

This gives us

${e^{ - {x^2}/2}}y' - {e^{ - {x^2}/2}}xy = {e^{ - {x^2}/2}}{x^2}$

$({e^{{x^2}/2}}y)' = {e^{{x^2}/2}}{x^2}$

Then we integrate to get

${e^{ - {x^2}/2}}y = \int {{e^{ - {x^2}/2}}{x^2}dx}$

We may try to solve the RHD integral by integration by parts. If we try directly we will find that we cannot integrate the exponential function, and integrating the x² part will just increase the power. Here we will use a little trick and split the function in another way to get

$\begin{gathered} I = \int {{e^{ - {x^2}/2}}{x^2}dx} = \int {\underbrace {{e^{ - {x^2}/2}}x}_f \cdot \underbrace x_gdx} = F \cdot g - \int {F \cdot g'dx} \hfill \\ \left| \begin{gathered} F = \int {{e^{ - {x^2}/2}}xdx} \hfill \\ \quad \left| {\begin{array}{*{20}{c}} {w = - {x^2}/2} \\ {dw = - xdx} \\ { - dw = xdx} \end{array}} \right. \hfill \\ = - \int {{e^w}dw = - {e^w} = - {e^{ - {x^2}/2}},\quad g' = 1} \hfill \\ \end{gathered} \right. \hfill \\ I = - {e^{ - {x^2}/2}}x + \int {{e^{ - {x^2}/2}}} dx \hfill \\ \end{gathered}$

And here we are stuck because the integral cannot be solved with elementary functions. So our answer, for the moment, is

${e^{ - {x^2}/2}}y = - {e^{ - {x^2}/2}}x + \int {{e^{ - {x^2}/2}}} dx$

$y = {e^{{x^2}/2}}\int {{e^{ - {x^2}/2}}} dx - x$

All is not lost though, because say we want to have a solution that goes through (1, 2) and we want to know its value when x=2. We could write our solution as

$y = {e^{{x^2}/2}}\int\limits_a^x {{e^{ - {x^2}/2}}dx} - x$

Where a is a constant. Next, we substitute our coordinates into the equation,

$2 = {e^{1/2}}\int\limits_a^1 {{e^{ - {x^2}/2}}dx} - 1$

$\frac{3}{{\sqrt e }} = \int\limits_a^1 {{e^{ - {x^2}/2}}dx}$

So we need to find a way to find this value a. How to do this? Yet again all is not lost. If you remember that the cumulative standard normal probability function is

$ncf(x) = \frac{1}{{\sqrt {2\pi } }}\int\limits_{ - \infty }^x {{e^{ - {t^2}/2}}dt}$

and most scientific calculators and mathematical software packages can calculate this (or something called the error function). So

$\int\limits_a^x {{e^{ - {x^2}/2}}dx} = \sqrt {2\pi } (ncf(x) - ncf(a))$

This, by the way, allows us to write the solution to our differential equation as

$y = {e^{{x^2}/2}}\sqrt {2\pi } (ncf(x) - ncf(a)) - x$

Ok, so we do have a general solution, but yet again, say we need a particular solution that passes through a point (x₀, y₀)? We get

$ncf(a) = ncf({x_0}) - \frac{{{x_0} + {y_0}}}{{{e^{{x_0}^2/2}}\sqrt {2\pi } }}$

or in our case (1, 2), so

$ncf(a) = ncf(1) - \frac{3}{{\sqrt {2\pi e} }}$

So we have the particular solution

$y = {e^{{x^2}/2}}\sqrt {2\pi } \left( {ncf(x) - ncf(1) + \frac{3}{{\sqrt {2\pi e} }}} \right) - x$

Back to the question, what is y when x=2? Substituting 2 into our equation we get about 13.96 for y.

A general solution that will go through a point (x₀, y₀) will thus be

$y = {e^{{x^2}/2}}\sqrt {2\pi } \left( {ncf(x) - ncf({x_0}) + \frac{{{x_0} + {y_0}}}{{{e^{{x_0}^2/2}}\sqrt {2\pi } }}} \right) - x$

This can be tested with this Texas instruments Nspire file: ypxyisx2
or this Geogebra file.

Up a level : Differential Equations
Previous page : Homogenous differential equations - a more general type
Next page : Brine mixture, case 1, inflow=outflow, pure water in

Last modified: Dec 28, 2023 @ 17:23