First order linear differential equations – a couple of examples

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Example 1

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}}}

Link to Geogebra file on this.

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