General Solution of Differential Equation

Show that the function

\[y(x) = \frac{C}{x^2} + \frac{3}{4}\]

is a solution of

\[2xy' = 3-4y\]

for any value of \(C\) which is a real number.

Solution:

First differentiating the function \(y(x)\) you get

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

Then substituting it into the left side of the equation,

\[ \begin{align} 2xy' &= 2x\left(-\frac{2C}{x^3} \right) \\ &= -\frac{4C}{x^2}. \end{align}\]

Substituting into the right side of the equation gives you

\[ \begin{align} 3-4y &= 3-4\left( \frac{C}{x^2} + \frac{3}{4} \right) \\ &= 3-\frac{4C}{x^2} - 3 \\ &=-\frac{4C}{x^2} .\end{align}\]

Since you get the same thing on the left and right sides when you substitute in \(y(x)\), it is a solution to the equation. In fact, this is true for any real number \(C\).

If you graph the solution for some values of \(C\) you can see why the general solution is often called a family of functions. The general solution defines an entire group of functions that are all very similar! All of the functions in the graph below have the same vertical asymptote, the same shape, and the same long-term behavior.

Fig. 2 - The general solution is a family of functions. Here you see four different values of \(C\) producing very similar looking curves.

What is the general solution to the homogeneous differential equation \(xy' = -2y \) ?

Solution:

This is a separable differential equation. It can be rewritten as

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

You can use an integrating factor to solve this, and for a reminder on how to do so see the article Solutions to Differential Equations. When you solve it you get

\[ y(x) = \frac{C}{x^2}.\]

Since the solution depends on a constant, it is a general solution. In fact, you could write it as

\[ y_C(x) = \frac{C}{x^2}.\]

to remind yourself that the general solution depends on that constant as well as on \(x\).

Show that \(y_p(x) = \dfrac{3}{4} \) solves the nonhomogeneous differential equation \(2xy' = 3-4y \).

Solution:

Notice that \(y'_p(x) = 0 \), so substituting this into the left side of the equation gives you

\[ 2xy_p' = 2x(0) = 0.\]

Substituting it into the right side of the equation,

\[ 3-4y_p = 3-4\left(\frac{3}{4}\right) = 0.\]

Since you get the same thing on both sides, \(y_p(x)\) is a solution to the nonhomogeneous differential equation.

Does \(y_C(x) = \dfrac{C}{x^2} \) solve the nonhomogeneous differential equation \(2xy' = 3-4y \)?

Solution:

Start by taking the derivative:

\[y'_C(x) = -\frac{2C}{x^3}.\]

Then substituting it into the differential equation on the left-hand side, you get

\[ \begin{align} 2xy_C' &= 2x\left( -\frac{2C}{x^3} \right) \\ &= -\frac{4C}{x^2} ,\end{align}\]

and on the right-hand side, you get

\[\begin{align} 3-4y_C &= 3-4\left(\frac{C}{x^2} \right) \\ &= 3-\frac{4C}{x^2} .\end{align}\]

These are definitely not the same, so \(y_C(x)\) does not solve the nonhomogeneous differential equation.

Show that \(y_C(x) = \dfrac{C}{x^2} \) solves the corresponding homogeneous differential equation \(2xy' = -4y \).

Solution:

As before,

\[y'_C(x) = -\frac{2C}{x^3},\]

and substituting this into the left side of the equation still gives you

\[ 2xy_C' = -\frac{4C}{x^2} .\]

However, substituting \(y_C(x)\) into the right side of the equation now gives you

\[ -4y_C = -\frac{4C}{x^2} ,\]

as well, so \(y_C(x)\) solves the corresponding homogeneous differential equation.

Which of the following is a general solution to the nonhomogeneous differential equation

\[y' = y+\sin x?\]

(a) \(y(x) = Ce^x\)

(b) \(y(x) = \sin x + \cos x\)

(c) \(y(x) = Ce^x -\dfrac{1}{2}(\sin x + \cos x )\).

Solution:

To figure this out, you can either solve the nonhomogeneous differential equation, or you can try plugging each one in. As you practice more you will get used to looking at an equation and having a general idea of what the solution will be. Let's look at each one of the potential solutions in turn.

(a) From experience working with linear differential equations you already know that \(y(x) = Ce^x\) is the solution to the homogeneous differential equation \(y'=y\). This is the general solution to the corresponding homogeneous differential equation of the nonhomogeneous differential equation. In other words, this would be \(y_C(x)\), and you have already seen that \(y_C(x)\) does not solve the nonhomogeneous differential equation.

(b) This potential solution looks more promising since it has trigonometric functions in it. If you plug it into the right-hand side of the nonhomogeneous differential equation you get

\[ \begin{align} y+\sin x &= \sin x + \cos x + \sin x \\ &= 2\sin x + \cos x. \end{align}\]

Taking the derivative you get

\[y'(x) = \cos x -\sin x.\]

Not quite the same, so this function is not the general solution to the nonhomogeneous differential equation.

(c) This potential solution has both the solution to the corresponding homogeneous differential equation and trigonometric functions. It might work! Taking the derivative you get

\[y'(x) = Ce^x -\frac{1}{2}(\cos x - \sin x).\]

Plugging it into the right-hand side of the equation you get

\[ \begin{align} y+\sin x &= Ce^x -\frac{1}{2}(\sin x + \cos x ) + \sin x \\ &= Ce^x +\frac{1}{2}\sin x -\frac{1}{2} \cos x \\ &= Ce^x -\frac{1}{2}(\cos x - \sin x) .\end{align}\]

Since you get the same thing on both sides, this function is a general solution to the nonhomogeneous differential equation.