Solving Simple Differential Equations

Solving differential equations of the form $\frac{dy}{dx} = f(x)$

Solving the expression $\frac{dy}{dx} = f(x)$ is really asking you to find what y is, expressed in terms of x, such that when you differentiate y you get f(x). This is a bit of a mouthful. Another way of saying this is: what do you differentiate to get f(x)?

We will look at a couple of examples, firstly without any additional conditions and then with additional conditions [you will see what we mean by additional conditions below]:

Example

Solve $\frac{dy}{dx} = 3x^2 + 4x - 3$

To find y we integrate both sides with respect to [wrt] x. When we integrate $\frac{dy}{dx}$ wrt x we get y [see Fundamental Theorem of Calculus above]. And when we integrate $3x^2 + 4x - 3$ wrt x we get $x^3 + 2x^2 - 3x + c$. So, the solution is

$y = x^3 + 2x^2 - 3x + c$

A few things to note in this example: when integrating $\frac{dy}{dx}$ we found $\int \frac{dy}{dx} dx$ and we used the Fundamental Theorem of Calculus which tells us that integration and differentiation are linked. So, if we differentiate y and then integrate it we should get just y back. Well actually, we should write y + constant but we don't as we tend to assume all the constants generated from integration are combined on the x side of the solution and written as "c".

We can now look at an example with some additional condition attached to it:

Example

Given $y = 5$ when $x = 1$ and $\frac{\mathrm{d}y}{\mathrm{d}x} = 3x^2 + 4x - 3$, find $y$ in terms of $x$

[Numerical values such as $y = 5$ when $x = 1$ are sometimes called "boundary conditions" or "initial conditions" depending on the circumstances and values given. We won't use these terms in the TMUA/ESAT.]

We can solve this two [entirely equivalent] ways.

Method 1

We can work out the answer as we did above with a constant term $c$ in it and then use the condition $y = 5$ when $x = 1$ to find the value of $c$ by substitution:

$y = x^3 + 2x^2 - 3x + c$

So $y = 5$ when $x = 1$ gives

$5 = 1 + 2 - 3 + c$

So $c = 5$

And the solution is

$y = x^3 + 2x^2 - 3x + 5$

Method 2

We notice that $y = 5$ when $x = 1$ and $y$ is $y$ when $x$ is $x$, and so we put these values directly as corresponding limits [top limits go together, and bottom limits go together] in the integration:

$\int_{y=5}^{y=y} \frac{\mathrm{d}y}{\mathrm{d}x} \mathrm{d}x = \int_{x=1}^{x=x} 3x^2 + 4x - 3 \mathrm{d}x$

Don't worry too much about the expression on the left [but do think a little about what it means]. (We have been a little slapdash with our notation on the left-hand side: as we are integrating wrt $x$ then the limits ought to be $x$ values and not the corresponding $y$ values, but it is easier just to put the $y$ values in as those are the ones we will end up using when we have integrated.) This leads to:

$[y]_5^y = [x^3 + 2x^2 - 3x]_1^x$

Which gives:

$y - 5 = (x^3 + 2x^2 - 3x) - (1 + 2 - 3)$

Which gives, as above:

$y = x^3 + 2x^2 - 3x + 5$

Much of this section has been covered earlier in these notes. And so, we will concentrate mainly on the parts of the specification that have not been covered in detail elsewhere and add a few additional comments when useful.