Particular Solutions to Differential Equations

Consider the linear differential equation initial value problem

\[ \begin{align} &y' -\frac{y}{x} = 3x \\ & y(1) = 7 .\end{align}\]

First, find the general solution, then find the particular solution if possible.

Solution:

First, let's solve the differential equation to get the general solution. Here \(P(x) = -1/x\) and \(Q(x) = 3x\), so you know the integrating factor is

\[ \begin{align} \exp\left( -\int \frac{1}{x} \, \mathrm{d} x\right) &= \exp\left(-\log x\right) = \frac{1}{x}.\end{align} \]

That means the solution to

\[ y' -\frac{y}{x} = 3x \]

is given by

\[ \begin{align} y\left(\frac{1}{x}\right) &= \int 3x\left(\frac{1}{x}\right)\, \mathrm{d}x \\ &= \int 3 \, \mathrm{d}x \\ &= 3x + C. \end{align}\]

Then solving for \(y\) you get

\[ y(x) = 3x^2 + Cx.\]

So the general solution is \(y(x) = 3x^2 + Cx \).

The particular solution makes use of the initial values to figure out what \(C\) is. Here the initial value is \(y(1) = 7\). Plugging that into the general solution you get

\[ 7 = 3(1)^2 + C\cdot 1,\]

or

\[ 4 = C.\]

So the particular solution to the initial value problem is

\[ y(x) = 3x^2 + 4x.\]

Let's go back to the linear differential equation, but with a different initial value. Is there a particular solution to

\[ \begin{align} &y' -\frac{y}{x} = 3x \\ & y(0) = 7 \end{align}\]

Solution:

From the previous example, you know that the general solution to

\[ y' -\frac{y}{x} = 3x \]

is

\[ y(x) = 3x^2 + Cx.\]

Now try plugging in the initial value to find \(C\). When you do,

you get

\[ 7 = 3(0)^2 + C\cdot 0,\]

or

\[ 7 = 0.\]

Hey, wait a minute! Seven doesn't equal zero, so what gives? Since you can't find a \(C\) that satisfies the initial value, this initial value problem doesn't have a particular solution!

Let's go back to the linear differential equation, but with a different initial value. Is there a particular solution to

\[ \begin{align} &y' -\frac{y}{x} = 3x \\ & y(0) = 0 \end{align}\]

Solution:

From the previous example you know that the general solution to

\[ y' -\frac{y}{x} = 3x \]

is

\[ y(x) = 3x^2 + Cx.\]

Now try plugging in the initial value to find \(C\). When you do,

you get

\[ 0 = 3(0)^2 + C\cdot 0,\]

or

\[ 0= 0.\]

Hey, wait a minute, that is always true! It doesn't matter what value of \(C\) you put in, it will always satisfy the initial value. That means this initial value problem has infinitely many solutions!

Find the particular solution to the initial value problem

\[ \begin{align} & y' = \dfrac{y^2}{x} \\ & y(1) = 2 \end{align}\]

along with any domain restrictions it might have.

Solution:

First let's find the solution. Separate the variables to get

\[ \frac{1}{y^2} y' = \frac{1}{x} \]

and then integrate both sides with respect to \(x\) to get

\[ \int \frac{1}{y^2} \, \mathrm{d} y = \int \frac{1}{x} \, \mathrm{d} x \]

so

\[ -\frac{1}{y} = \ln |x| + C.\]

Then solving for \(y\), the general solution is given by

\[ y(x) = -\frac{1}{ \ln |x| + C }.\]

Now you can use the initial condition \(y(1)=2\) to find a particular solution. That means

\[ 2 = -\frac{1}{ \ln |1| + C },\]

and

\[C = -\frac{1}{2}.\]

So the particular solution is

\[ y(x) = -\frac{1}{ \ln |x| - \frac{1}{2} }.\]

Now let's look at any restrictions that might be on the solution. With the absolute value signs there, you don't need to worry about taking the log of a negative number. However you still can't have \(x=0\), and you also need that the denominator is not zero. That means you need

\[ \ln |x| - \frac{1}{2} \ne 0.\]

Using properties of logarithms, you can see that \(x \ne \pm \sqrt{e}\) is also a necessary condition.

That means there are four intervals that your solution might be in:

• \( -\infty < x < -\sqrt{e} \)
• \( -\sqrt{e} < x < 0 \)
• \(0 < x < \sqrt{e}\)
• \( \sqrt{e} < x < \infty\).

So how do you know which one your solution is in? Just look at the initial value! The initial value for this problem is \(y(1) = 2 \), and \(x=1\) is in the interval \( (0 , \sqrt{e} )\). That means the domain restriction for this particular solution is \( (0 , \sqrt{e} )\).

Show that

\[ y = 2x^{-3}\]

is a particular solution of the initial value problem

\[ \begin{align} &xy' +3y = 0 \\ &y(1) = 2. \end{align}\]

Solution:

It is usually a good idea to check the initial value first since it will be relatively easy, and if the prospect doesn't satisfy the initial value it can't be a solution to the initial value problem. In this case,

\[ \begin{align} y(1) & = 2(1)^{-3} \\ &= 2, \end{align}\]

so the function \(y(x) = 2x^{-3} \) does satisfy the initial value. Now you just need to check to see if it satisfies the equation. For that you need \(y'\), so

\[ y' = 2(-3)(x^{-4}) = -6x^{-4}.\]

Substituting that into the differential equation,

\[ \begin{align} xy' +3y &= x\left(-6x^{-4} \right) + 3\left(2x^{-3} \right) \\ &= -6x^{-3} + 6x^{-3} \\ &= 0 \end{align}\]

So the proposed solution does satisfy the differential equation.

Since \(y(x) = 2x^{-3} \) satisfies both the initial value and the differential equation, it is a particular solution to the initial value problem.

Find a particular solution to the initial value problem

\[ \begin{align} &y'' = 3x+2 \\ &y(0)=3 \\ &y'(0) = 1. \end{align}\]

Solution:

The first step is to find a general solution. Notice that this is actually a second-order equation, so it has two initial values. However this is an especially nice second-order equation since the only \(y\) in it is a second derivative, and it is already separated.

Integrating both sides of the equation with respect to \(x\) you get

\[ y' = \frac{3}{2}x^2 + 2x + C.\]

Integrating once more you get

\[ y(x) = \frac{1}{2}x^3 + x^2 + Cx + D,\]

which is the general solution. There are two constants to go with the two initial values. Using \(y'(0) = 1 \) you get

\[ y'(0) = \frac{3}{2}0^2 + 2(0) + C = 1,\]

So \(C = 1\). Plugging that in to the general solution gives you

\[ y(x) = \frac{1}{2}x^3 + x^2 + x + D,\] and then you can use the second initial value \(y(0)=3 \) to get

\[ y(0) = \frac{1}{2}0^3 + 0^2 +0 + D = 3,\]

which means that \(D = 3\). Therefore the particular solution to the initial value problem is

\[ y(x) = \frac{1}{2}x^3 + x^2 + x + 3.\]