General and Particular Solutions

What is a Solution to a Differential Equation?

In an algebraic equation such as $x^2 - 5x + 6 = 0$, the solution is a number (or numbers) that satisfies the equation.

In a differential equation, the solution is usually a function or a relation between the variables that satisfies the equation. If we substitute that function and all the required derivatives into the differential equation, the left-hand side and right-hand side must become identical.

For example, consider $$\frac{dy}{dx} = 2x.$$ The function $$y = x^2$$ is a solution because $$\frac{dy}{dx} = \frac{d}{dx}(x^2) = 2x,$$ which exactly matches the differential equation.

So, a solution of a differential equation is not merely a number; it is generally a curve, a family of curves, or an implicit relation that satisfies the equation.

The General Solution

For the differential equation $$\frac{dy}{dx} = 2x,$$ not only $y = x^2$ but also $$y = x^2 + 5, \quad y = x^2 - 10, \quad y = x^2 + C$$ are all valid solutions.

The solution that contains one or more arbitrary constants is called the general solution (or primitive) of the differential equation.

Key Rule: The number of arbitrary constants in the general solution is equal to the order of the differential equation.

• A first-order differential equation has a general solution containing one arbitrary constant.
• A second-order differential equation has a general solution containing two arbitrary constants.
• In general, an $n$th order differential equation has a general solution containing $n$ arbitrary constants.

Geometrically, the general solution represents a family of curves.

The Particular Solution

A particular solution is obtained from the general solution by assigning specific values to all arbitrary constants.

These specific values are usually found using extra conditions such as:

• initial conditions, for example $y=5$ when $x=0$,
• boundary conditions, for example values of the function at two given points.

For example, if the general solution is $$y = x^2 + C,$$ and we are given the condition $y=3$ when $x=1$, then $$3 = 1^2 + C \implies C = 2.$$ So the particular solution becomes $$y = x^2 + 2.$$

Key Rule: A particular solution contains no arbitrary constants.

Geometrically, a particular solution represents one specific curve chosen from the entire family of curves represented by the general solution.

Example 1: Verifying a Particular Solution (First Order)

Verify that the function $y = e^{-3x}$ is a solution of the differential equation $\frac{dy}{dx} + 3y = 0$.

Solution: Step 1: Identify the given function and the differential equation. Given function: $$y = e^{-3x}$$ Differential equation: $$\frac{dy}{dx} + 3y = 0.$$

Step 2: Differentiate the given function. Using the chain rule, $$\frac{dy}{dx} = \frac{d}{dx}(e^{-3x}) = -3e^{-3x}.$$

Step 3: Substitute $y$ and $\frac{dy}{dx}$ into the left-hand side of the differential equation. $$\frac{dy}{dx} + 3y = (-3e^{-3x}) + 3(e^{-3x}).$$

Step 4: Simplify. $$-3e^{-3x} + 3e^{-3x} = 0.$$ So the left-hand side becomes 0, which is equal to the right-hand side.

Step 5: Conclusion. Hence, $$y = e^{-3x}$$ is a valid solution of the differential equation.

Since it contains no arbitrary constant, it is a particular solution, not a general solution.

Example 2: Verifying a General Solution (Second Order)

Verify that the function $y = A \cos x + B \sin x$ is a solution of the differential equation $\frac{d^2y}{dx^2} + y = 0$.

Solution: Step 1: Write the given function. $$y = A\cos x + B\sin x$$ where $A$ and $B$ are arbitrary constants.

Step 2: Find the first derivative. $$\frac{dy}{dx} = -A\sin x + B\cos x.$$

Step 3: Find the second derivative. $$\frac{d^2y}{dx^2} = -A\cos x - B\sin x.$$

Step 4: Substitute into the left-hand side of the differential equation. $$\frac{d^2y}{dx^2} + y = (-A\cos x - B\sin x) + (A\cos x + B\sin x).$$

Step 5: Simplify. $$-A\cos x + A\cos x - B\sin x + B\sin x = 0.$$ Thus the left-hand side becomes 0, which matches the right-hand side.

Step 6: Conclusion. Therefore, $$y = A\cos x + B\sin x$$ is a solution of the differential equation.

Because the differential equation is of order 2 and the solution contains two arbitrary constants $A$ and $B$, it is the general solution.

Example 3: Verifying a Particular Solution

Verify that $y = \sqrt{1 + x^2}$ is a solution of the differential equation $y' = \frac{x}{y}$.

Solution: Step 1: Write the given function in power form. $$y = \sqrt{1+x^2} = (1+x^2)^{1/2}.$$

Step 2: Differentiate the function. Using the chain rule, $$y' = \frac{1}{2}(1+x^2)^{-1/2}\cdot 2x = \frac{x}{\sqrt{1+x^2}}.$$

Step 3: Express the derivative in terms of $y$. Since $$y = \sqrt{1+x^2},$$ we get $$y' = \frac{x}{y}.$$

Step 4: Compare with the given differential equation. The required differential equation is exactly $$y' = \frac{x}{y}.$$ Hence the given function satisfies the differential equation.

Step 5: Conclusion. Therefore, $$y = \sqrt{1+x^2}$$ is a valid solution. Since there is no arbitrary constant, it is a particular solution.

Example 4: Number of Arbitrary Constants

Determine the number of arbitrary constants in the general solution and the particular solution of a 4th order differential equation.

Solution: Step 1: Use the rule for the general solution. The number of arbitrary constants in the general solution is equal to the order of the differential equation. Since the equation is of order 4, its general solution contains 4 arbitrary constants.

Step 2: Use the rule for the particular solution. A particular solution is obtained after assigning specific values to all arbitrary constants. Therefore, a particular solution contains 0 arbitrary constants.

Answer:

• General solution: 4 arbitrary constants
• Particular solution: 0 arbitrary constants

Example 5: Verifying an Implicit Solution

Verify that the relation $x + y = \tan^{-1} y$ is a solution of the differential equation $y^2 y' + y^2 + 1 = 0$.

Solution: Step 1: Differentiate the given implicit relation with respect to $x$. Given, $$x + y = \tan^{-1} y.$$ Differentiating both sides with respect to $x$, $$\frac{d}{dx}(x+y) = \frac{d}{dx}(\tan^{-1}y).$$ This gives $$1 + y' = \frac{1}{1+y^2} \cdot y'.$$

Step 2: Rearrange to collect the $y'$ terms. $$1 = \frac{y'}{1+y^2} - y'.$$ Take $y'$ common: $$1 = y'\left(\frac{1}{1+y^2} - 1\right).$$

Step 3: Simplify the bracket. $$\frac{1}{1+y^2} - 1 = \frac{1-(1+y^2)}{1+y^2} = \frac{-y^2}{1+y^2}.$$ So, $$1 = y'\left(\frac{-y^2}{1+y^2}\right).$$

Step 4: Multiply through by $(1+y^2)$. $$1+y^2 = -y^2 y'.$$ Bring all terms to one side: $$y^2 y' + y^2 + 1 = 0.$$

Step 5: Conclusion. This is exactly the given differential equation. Therefore, $$x+y = \tan^{-1}y$$ is a valid implicit solution.

Example 6: Solution Verification with Parameters

Verify that $y = ax + \frac{b}{x}$ is a solution of the differential equation $x^2 \frac{d^2y}{dx^2} + x \frac{dy}{dx} - y = 0$.

Solution: Step 1: Write the given function in index form. $$y = ax + bx^{-1}.$$

Step 2: Find the first derivative. $$y' = a - bx^{-2} = a - \frac{b}{x^2}.$$

Step 3: Find the second derivative. $$y'' = 0 + 2bx^{-3} = \frac{2b}{x^3}.$$

Step 4: Substitute $y$, $y'$, and $y''$ into the left-hand side. $$x^2y'' + xy' - y = x^2\left(\frac{2b}{x^3}\right) + x\left(a - \frac{b}{x^2}\right) - \left(ax + \frac{b}{x}\right).$$

Step 5: Simplify each term. $$x^2\left(\frac{2b}{x^3}\right) = \frac{2b}{x},$$ $$x\left(a - \frac{b}{x^2}\right) = ax - \frac{b}{x}.$$ So, $$x^2y'' + xy' - y = \frac{2b}{x} + ax - \frac{b}{x} - ax - \frac{b}{x}.$$

Step 6: Combine like terms. $$ax-ax = 0, \qquad \frac{2b}{x} - \frac{b}{x} - \frac{b}{x} = 0.$$ Hence, $$x^2y'' + xy' - y = 0.$$

Step 7: Conclusion. Therefore, $$y = ax + \frac{b}{x}$$ is a solution of the differential equation. Since it contains two arbitrary constants $a$ and $b$, it represents a general solution candidate for this second-order equation.

Example 7: Exponential Solution

Verify that $y = C_1 e^{2x} + C_2 e^{-2x}$ is a solution of $y'' - 4y = 0$.

Solution: Step 1: Find the first derivative. $$y' = 2C_1 e^{2x} - 2C_2 e^{-2x}.$$

Step 2: Find the second derivative. $$y'' = 4C_1 e^{2x} + 4C_2 e^{-2x}.$$

Step 3: Factor out 4 from the second derivative. $$y'' = 4\left(C_1 e^{2x} + C_2 e^{-2x}\right).$$ But the bracket is exactly $y$. Therefore, $$y'' = 4y.$$

Step 4: Rearrange. $$y'' - 4y = 0.$$

Step 5: Conclusion. Hence, $$y = C_1 e^{2x} + C_2 e^{-2x}$$ is a valid solution of the differential equation. Since the differential equation is second order and the solution contains two arbitrary constants, it is the general solution.