Differential Equations

Video lesson · from 15:162 worked examples

One lesson video covers all of Integration, so it opens at 15:16 for this topic — not from the beginning.

Theory

As previously stated, integration is the reverse of differentiation.

As a result, if we integrate the derivative of a function then we obtain the original function:

\int \frac{dy}{dx} \,dx = y

\int f'(x) \,dx = f(x)

The above obtains a general solution for the original function.

However, if we have additional information about the function, we can find the value of the constant of integration ( $C$ ) to obtain a particular solution.

⚠️ Common Examiner Traps

You cannot skip the constant here: the whole point is to find it. Integrate, write $+\,C$ , then use the given condition to pin it down.
Substitute the condition immediately: put the given point into your integrated expression before doing anything else, and solve for $C$ .
Write the final function out: the answer is the equation with $C$ replaced by its value — not the value of $C$ on its own.
Read what the derivative represents: in a context question, $\frac{dV}{dt}$ is a rate. Integrating gives the quantity, not the rate.

Worked examples

Example 1

The graph of $y=f(x)$ passes through the point $(3, -4)$ .
If $\frac{dy}{dx} = x^2 - 5$ , express $y$ in terms of $x$ .

\begin{aligned} y &= \int (x^2 - 5) \,dx \\ &= \frac{x^3}{3} - 5x + C \end{aligned}

Substitute $x=3, y=-4$ :

\begin{aligned} -4 &= \frac{3^3}{3} - 5(3) + C \\ -4 &= 9 - 15 + C \\ -4 &= -6 + C \\ 2 &= C \end{aligned}

Solution: $y = \frac{x^3}{3} - 5x + 2$

Example 2

The function is defined on a suitable domain such that $f'(x) = x^2 + \frac{1}{x^2} + \frac{2}{3}$ .
Given that $f(1) = 4$ , find $f(x)$ in terms of $x$ .

\begin{aligned} f(x) &= \int (x^2 + x^{-2} + \frac{2}{3}) \,dx \\ &= \frac{x^3}{3} - x^{-1} + \frac{2}{3}x + C \\ &= \frac{x^3}{3} - \frac{1}{x} + \frac{2}{3}x + C \end{aligned}

Substitute $x=1, f(1)=4$ :

\begin{aligned} 4 &= \frac{1^3}{3} - \frac{1}{1} + \frac{2}{3}(1) + C \\ 4 &= \frac{1}{3} - 1 + \frac{2}{3} + C \\ 4 &= C \end{aligned}

Solution: $f(x) = \frac{x^3}{3} - \frac{1}{x} + \frac{2}{3}x + 4$