Solution To A Differential Equation

Solutions of differential equations

A solution to a differential equation is a function $y(x)$ whose derivatives $y'(x), y''(x), ..., y^{(n)}(x)$ exist and satisfy the differential equation in the interval where the equation is defined.

Verification of a solution of a differential equation

To verify if a function $y = f(x)$ is a solution to a given differential equation:

1. Identify the function and the differential equation.
2. Find the required derivatives of the given function.
3. Substitute the derivatives and the original function into the differential equation.
   • If a true statement is obtained (LHS = RHS), the function is a solution.
   • If a false statement is obtained (LHS ≠ RHS), the function is not a solution.

Example 1

Verify that $f(x) = 2\sin x + 2\cos x$ is a solution of the differential equation $f''(x) + f(x) = 0$.

Solution:

Given $f(x) = 2\sin x + 2\cos x$.

$f'(x) = 2\cos x - 2\sin x$

$f''(x) = -2\sin x - 2\cos x$

Substituting into the differential equation:

$f''(x) + f(x) = (-2\sin x - 2\cos x) + (2\sin x + 2\cos x) = 0$

The equation is satisfied. Therefore, $f(x) = 2\sin x + 2\cos x$ is a solution.

Example 2

Show that $x^2 + y^2 - 16 = 0$ is a solution of the differential equation $xy' + y = 0$ in the interval $(-4, 4)$.

Solution:

Given $x^2 + y^2 - 16 = 0$.

Differentiating implicitly with respect to $x$:

$2x + 2y\frac{dy}{dx} = 0$

$x + y\frac{dy}{dx} = 0$

$xy' + y = 0$

The given equation is satisfied. Therefore, $x^2 + y^2 - 16 = 0$ is a solution.

Alternatively, solving for $y$: $y = \pm\sqrt{16-x^2}$. Then $y' = \mp \frac{x}{\sqrt{16-x^2}}$. Substituting into $xy' + y$:

$x\left(\mp \frac{x}{\sqrt{16-x^2}}\right) \pm \sqrt{16-x^2} = \mp\frac{x^2}{\sqrt{16-x^2}} \pm \frac{16-x^2}{\sqrt{16-x^2}} = \frac{\mp x^2 \pm (16-x^2)}{\sqrt{16-x^2}} = \frac{\pm 16 \mp 2x^2}{\sqrt{16-x^2}}$

This is only equal to zero if we consider the implicit differentiation result. The previous simplification in the original text was incorrect.

Example 3

Verify that $x(\theta) = \frac{1}{6}e^{3\theta} + c_1e^{\theta} + c_2e^{-\theta}$ is a solution of the differential equation $x''(\theta) - x'(\theta) - 2x(\theta) = e^{3\theta}$.

Solution:

Given $x(\theta) = \frac{1}{6}e^{3\theta} + c_1e^{\theta} + c_2e^{-\theta}$.

$x'(\theta) = \frac{1}{2}e^{3\theta} + c_1e^{\theta} - c_2e^{-\theta}$

$x''(\theta) = \frac{3}{2}e^{3\theta} + c_1e^{\theta} + c_2e^{-\theta}$

Substituting into the differential equation:

$x''(\theta) - x'(\theta) - 2x(\theta) = \left(\frac{3}{2}e^{3\theta} + c_1e^{\theta} + c_2e^{-\theta}\right) - \left(\frac{1}{2}e^{3\theta} + c_1e^{\theta} - c_2e^{-\theta}\right) - 2\left(\frac{1}{6}e^{3\theta} + c_1e^{\theta} + c_2e^{-\theta}\right)$

$= \frac{3}{2}e^{3\theta} - \frac{1}{2}e^{3\theta} - \frac{1}{3}e^{3\theta} + c_1e^{\theta} - c_1e^{\theta} - 2c_1e^{\theta} + c_2e^{-\theta} + c_2e^{-\theta} - 2c_2e^{-\theta}$

$= \left(\frac{9-3-2}{6}\right)e^{3\theta} + (1-1-2)c_1e^\theta + (1+1-2)c_2e^{-\theta}$

$= e^{3\theta}$

The equation is satisfied. Therefore, the given function is a solution.