Recognising an exact equation

5 Recognising an exact equation

The equation $\frac{d}{d x} (y x) = 3 x^{2}$ is exact, as we have seen. If we expand the left-hand side of this equation (i.e. differentiate the product) we obtain

$x \frac{d y}{d x} + y .$

Hence the equation

$x \frac{d y}{d x} + y = 3 x^{2}$

must be exact, but it is not so obvious that it is exact as in the original form. This leads to the following Key Point:

Key Point 3

The equation

f (x) \frac{d y}{d x} + y f^{'} (x) = g (x)

is exact. It can be re-written as

\frac{d}{d x} (y f (x)) = g (x) so that y f (x) = \int g (x) d x

Example 4

Solve the equation

$x^{3} \frac{d y}{d x} + 3 x^{2} y = x$

Solution

Comparing this equation with the form in Key Point 3 we see that $f (x) = x^{3}$ and $g (x) = x$ . Hence the equation can be written

$\frac{d}{d x} (y x^{3}) = x$

which has solution

$y x^{3} = \int x d x = \frac{1}{2} x^{2} + C .$

Therefore

$y = \frac{1}{2 x} + \frac{C}{x^{3}} .$

Task!

Solve the equation $sin x \frac{d y}{d x} + y cos x = cos x$ .

You should obtain $y = 1 + C cosec x$ since, here $f (x) = sin x$ and $g (x) = cos x$ . Then

$\frac{d}{d x} (y sin x) = cos x and y sin x = \int cos x d x = sin x + C$

Finally $y = 1 + C cosec x$ .

Exercises

Solve the equation $\frac{d}{d x} (y x^{2}) = x^{3} .$
Solve the equation $\frac{d}{d x} (y e^{x}) = e^{2 x}$ given the condition $y (0) = 2.$
Solve the equation $e^{2 x} \frac{d y}{d x} + 2 e^{2 x} y = x^{2}$ .
Show that the equation $x^{2} \frac{d y}{d x} + 2 x y = x^{3}$ is exact and obtain its solution.
Show that the equation $x^{2} \frac{d y}{d x} + 3 x y = x^{3}$ is not exact.
Multiply the equation by $x$ and show that the resulting equation is exact and obtain its solution.

$y = \frac{x^{2}}{4} + \frac{C}{x^{2}}$ .
$y = \frac{1}{2} e^{x} + \frac{3}{2} e^{- x}$ .
$y = (\frac{1}{3} x^{3} + C) e^{- 2 x}$ .
$y = \frac{1}{4} x^{2} + \frac{C}{x^{2}}$ .
$y = \frac{1}{5} x^{2} + \frac{C}{x^{3}}$ .