Separable Equations

Identify the following first order equations as separable or non-separable.

(a) \(y'=(x^{2}+9)5y \)

(b) \(y'=3x^{2}-4x \)

(c) \(\ln(y') = x+5\)

(d) \(y'=xy+3x-2y-6 \)

(e) \(e^{y'} = x + y\)

Answer:

(a) If you let \(f(x) = x^{2}+9 \) and \(g(y) = 5y \), then you can write \(y' = f(x)g(y)\), so this is a separable equation.

(b) This one is already separated, although it might take you a moment to notice that you can let \(g(y) = 1\).

There is no reason that you can't let \(f(x)\) or \(g(y)\) be constant functions!

(c) The equation might not look like it is separable, but let's use properties of logarithms. Remember that \(\ln(y') = x+5\) means the same thing as

\[ y' = e^{x+5}.\]

So this is a separable equation.

(d) The differential equation \(y'=xy+3x-2y-6 \) doesn't look separable at first glance, but let's do some algebra just to be sure. If you factor the right hand side, you get

\[ \begin{align} y' &= xy+3x-2y-6 \\ \\ &= x(y+3) - 2(y+3) \\ \\ &= (x-2)(y+3). \end{align}\]

So in fact this equation is separable.

(e) Let's try using properties of logarithms on this example. If you do, \(e^{y'} = x + y\) becomes

\[ \ln e^{y'} = \ln (x+y), \]

or in other words

\[ y' = \ln (x+y) .\]

There is no way to write the right-hand side of that equation as \(f(x)g(y)\) because you have the \(x+y\) term inside the logarithm. Therefore this equation is not separable.

For the differential equation is \(y' = (x-2)(y+3) \), find any constant solutions.

Answer:

Constant solutions happen when \(g(y) = 0\). For this problem, \(g(y) = y + 3\), and this equals zero when \(y = -3\). So there is a constant solution, and it is \(y=-3\).

Solve the differential equation

\[\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{3x^{2}}{\cos y}.\]

Answer:

Always check first to see that the equation is separable!

To make sure the differential equation is separable, you have to write the equation in the form \(y'=f(x)g(y)\).

If you let \(f(x)=3x^{2}\) and \(g(y)=1/ \cos y\), then you can see that the equation is, in fact, separable.

It is always a good idea to look for places where \(g(y) = 0\) since these correspond to constant solutions. In this case, \(g(y)\) can never be zero, though it can be undefined. So there won't be any constant solutions, but there might be places where the solution to the differential equation doesn't exist.

Next you want to set up the integral. Since

\[\frac{\mathrm{d}y}{\mathrm{d}x} =\frac{3x^{2}}{\cos y},\]

or

\[\cos y \frac{\mathrm{d}y}{\mathrm{d}x} =3x^{2},\]

you might be tempted to "cross multiply by \(\mathrm{d}x \)" and then integrate. You can't actually do that because \(\mathrm{d}y/ \mathrm{d}x\) isn't really a fraction. But the really cool thing is that you can pretend to do it, and it makes the integral

\[ \int \cos y \,\mathrm{d}y = \int 3x^2 \,\mathrm{d}x ,\]

which is exactly what you would have gotten by doing the \(u\) substitution!

Then integrating you get

\[ \sin y = x^3 + C.\]

Remember that this is called an implicit solution, since you don't have \(y\) by itself. To get an explicit solution, you can solve for \(y\) and see that

\[ y(x) = \arcsin(x^3 + C).\]

What about an interval of existence? At the start of the problem, you noticed that when \(\cos y = 0\) there would be problems, and you can see from the explicit form of the solution that there is an arcsine in it, which is only defined on the interval

\[ \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right].\]

So that is the maximum interval of existence for the solution to the differential equation.

Answer:

You have already checked in previous examples that this is a separable equation, and found that there is a constant solution \(y = -3\). So now let's find the rest of the solutions.

Setting up the integral,

\[ \int \frac{1}{y+3}\, \mathrm{d}y = \int x-2 \,\mathrm{d}x .\]

Then integrating gives you

\[ \ln|y+3| = \frac{1}{2}x^2 - 2x + C \]

as the implicit solution. Before finding an explicit solution, notice that you need \(y \ne -3\) for this solution to be valid, since otherwise you would be trying to take the logarithm of zero!

Going on to find the explicit solution, using properties of logarithms,

\[ |y+3| = \exp \left( \frac{1}{2}x^2 - 2x + C \right) \]

where \( \exp\) is just a handy notation for the exponential function.

Now let \(A = e^C\) since that is just another constant. Then you can write

\[ |y+3| = A\exp \left( \frac{1}{2}x^2 - 2x \right),\]

but for it to be a true explicit solution you need to finish solving for \(y\). In fact, this corresponds to two different solutions,

\[ y =-3+ A\exp \left( \frac{1}{2}x^2 - 2x \right),\]

and

\[ y =-3- A\exp \left( \frac{1}{2}x^2 - 2x \right).\]

You can combine these into one by using the \(\pm\) sign to indicate that it could be positive or negative, to get

\[ y =-3 \pm A\exp \left( \frac{1}{2}x^2 - 2x \right).\]

So does the explicit solution also give you the constant solution, or does that need to be written separately? Notice that if \(A=0\) then in fact you do get \(y=-3\), so the explicit solution also covers the constant solution. That means the solution to the differential equation is

\[ y =-3 \pm A\exp \left( \frac{1}{2}x^2 - 2x \right).\]