Use separation of variables to solve a differential equation.
Solve applications using separation of variables.
We now examine a solution technique for finding exact solutions to a class of differential equations known as separable differential equations. These equations are common in a wide variety of disciplines, including physics, chemistry, and engineering. We illustrate a few applications at the end of the section.
Separation of variables
We start with a definition and some examples.
Definition
A
separable differential equation is any equation that can be written in the form
$y\prime =f\left(x\right)g\left(y\right).$
The term ‘separable’ refers to the fact that the right-hand side of the equation can be separated into a function of
$x$ times a function of
$y.$ Examples of separable differential equations include
The second equation is separable with
$f\left(x\right)=6{x}^{2}+4x$ and
$g\left(y\right)=1,$ the third equation is separable with
$f\left(x\right)=1$ and
$g\left(y\right)=\text{sec}\phantom{\rule{0.1em}{0ex}}y+\text{tan}\phantom{\rule{0.1em}{0ex}}y,$ and the right-hand side of the fourth equation can be factored as
$\left(x+3\right)\left(y-2\right),$ so it is separable as well. The third equation is also called an
autonomous differential equation because the right-hand side of the equation is a function of
$y$ alone. If a differential equation is separable, then it is possible to solve the equation using the method of
separation of variables .
Problem-solving strategy: separation of variables
Check for any values of
$y$ that make
$g(y)=0.$ These correspond to constant solutions.
Rewrite the differential equation in the form
$\frac{dy}{g(y)}=f(x)dx.$
Integrate both sides of the equation.
Solve the resulting equation for
$y$ if possible.
If an initial condition exists, substitute the appropriate values for
$x$ and
$y$ into the equation and solve for the constant.
Note that Step 4. states “Solve the resulting equation for
$y$ if possible.” It is not always possible to obtain
$y$ as an explicit function of
$x.$ Quite often we have to be satisfied with finding
$y$ as an implicit function of
$x.$
Using separation of variables
Find a general solution to the differential equation
$y\prime =\left({x}^{2}-4\right)\left(3y+2\right)$ using the method of separation of variables.
Follow the five-step method of separation of variables.
In this example,
$f\left(x\right)={x}^{2}-4$ and
$g\left(y\right)=3y+2.$ Setting
$g(y)=0$ gives
$y=-\frac{2}{3}$ as a constant solution.
To solve this equation for
$y,$ first multiply both sides of the equation by
$3.$
$\text{ln}\left|3y+2\right|={x}^{3}-12x+3C$
Now we use some logic in dealing with the constant
$C.$ Since
$C$ represents an arbitrary constant,
$3C$ also represents an arbitrary constant. If we call the second arbitrary constant
${C}_{1},$ the equation becomes
$\text{ln}\left|3y+2\right|={x}^{3}-12x+{C}_{1}.$
Now exponentiate both sides of the equation (i.e., make each side of the equation the exponent for the base
$e).$
Again define a new constant
${C}_{2}={e}^{{c}_{1}}$ (note that
${C}_{2}>0)\text{:}$
$\left|3y+2\right|={C}_{2}{e}^{{x}^{3}-12x}.$
This corresponds to two separate equations:
$3y+2={C}_{2}{e}^{{x}^{3}-12x}$ and
$3y+2=\text{\u2212}{C}_{2}{e}^{{x}^{3}-12x}.$ The solution to either equation can be written in the form
$y=\frac{\mathrm{-2}\pm {C}_{2}{e}^{{x}^{3}-12x}}{3}.$ Since
${C}_{2}>0,$ it does not matter whether we use plus or minus, so the constant can actually have either sign. Furthermore, the subscript on the constant
$C$ is entirely arbitrary, and can be dropped. Therefore the solution can be written as
$y=\frac{\mathrm{-2}+C{e}^{{x}^{3}-12x}}{3}.$
No initial condition is imposed, so we are finished.
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