# First order linear differential equations – a couple of examples

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Let us solve $y' - xy = x$

This is actually a separable equation, but let us first solve it as linear differential equation. We have p(x)= –x so that gives us $I = {e^{\int {( - x)dx} }} = {e^{ - \frac{{{x^2}}}{2}}}$

Multiplying our equation by this gives us ${e^{ - \frac{{{x^2}}}{2}}}y' - {e^{ - \frac{{{x^2}}}{2}}}xy = {e^{ - \frac{{{x^2}}}{2}}}x$

or ${\left( {{e^{ - \frac{{{x^2}}}{2}}}y} \right)^\prime } = {e^{ - \frac{{{x^2}}}{2}}}x$

So ${e^{ - \frac{{{x^2}}}{2}}}y = \int {{e^{ - \frac{{{x^2}}}{2}}}xdx}$

If we now do the variable substitution $\left| \begin{gathered} w = - \frac{{{x^2}}}{2} \hfill \\ - dw = xdx \hfill \\ \end{gathered} \right.$

We get ${e^{ - \frac{{{x^2}}}{2}}}y = - \int {{e^w}dw = {e^w} + C = - } {e^{ - \frac{{{x^2}}}{2}}} + C$

Multiplying away the first factor, we get $y = {e^{\frac{{{x^2}}}{2}}}( - {e^{ - \frac{{{x^2}}}{2}}} + C) = C{e^{\frac{{{x^2}}}{2}}} - 1$

that is our final solution. Interestingly enough it is the same solutions as we found to $\frac{{dy}}{{dx}} = xy$

but just shifted down one unit. As we can see we could write $y' - xy = x$

as $y' = x + xy$

or $y' = x(1 + y)$

so as mentioned it is actually separable. We get $\frac{{dy}}{{y + 1}} = xdx$

or $\ln |y + 1| = \frac{{{x^2}}}{2} + {C_1}$

and thus $y + 1 = C{e^{\frac{{{x^2}}}{2}}}$

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

just as expected.

Example 2

Let us take an old example and modify it slightly. If the slopes where directed out from the origin we had that $\frac{{dy}}{{dx}} = \frac{y}{x}$

and that gave us solutions of the form $y = Cx$

I.e. straight lines through the origin. Now let us change that a bit. Say we are subtracting or adding) a bit from the slope, and that what we subtract is proportional to the distance from the y-axis. Say we have $\frac{{dy}}{{dx}} = \frac{y}{x} - 0.1x$

This can be written as $y' - \frac{y}{x} = - 0.1x$

So it is a linear first order equation with $p(x) = - \frac{1}{x}$

This gives us $I = {e^{ - \int {\frac{{dx}}{x}dx} }} = {e^{ - \ln x}} = {e^{\ln {x^{ - 1}}}} = \frac{1}{x}$

Multiplying through with this gives us $\frac{{y'}}{x} - \frac{y}{{{x^2}}} = - 0.1$

or ${\left( {\frac{{y'}}{x}} \right)^\prime } = - 0.1$

Integration now gives us $\frac{{y}}{x} = - 0.1x + C$

that gives us $y = - 0.1{x^2} + Cx$

This looks rather pleasing as a graph, with a few particular solutions. As you can see all the solutions will be parabolas passing through the origin.

To find particular solutions passing through a particular point (x0,y0) we just substitute in those values into our solution to get ${y_0} = - 0.1x_0^2 + C{x_0}$

or $C = \frac{{{y_0} + 0.1x_0^2}}{{{x_0}}}$

This gives us the particular solution $C = \frac{{{y_0} + 0.1x_0^2}}{{{x_0}}}$ Up a level : Differential Equations
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Next page : Using Taylor series to solve differential equations Last modified: Dec 26, 2020 @ 15:13