Question

Find all functions \(\mathrm{f}(\mathrm{x})\) such that \(\mathrm{f}^{\prime \prime}(\mathrm{x})=2 \sin 3 \mathrm{x}\).

Short answer

Question: Find the general solution to the differential equation f''(x) = 2 sin(3x). Answer: The general solution to the given differential equation is f(x) = -2/9 sin(3x) + C_1x + C_2, where C_1 and C_2 are constants of integration.

Step-by-step solution

Integrate both sides once

We will integrate both sides of the equation f''(x) = 2 sin(3x) with respect to x: $$\int f''(x) dx = \int 2 \sin(3x) dx$$

Evaluate the integral on the right-hand side

To evaluate the integral on the right-hand side, we will use integration by substitution. Let u = 3x, so du/dx = 3 and dx = du/3: $$\int 2 \sin(3x) dx = \frac{2}{3} \int \sin(u) du$$ Now, integrate sin(u) with respect to u: $$\frac{2}{3} \int \sin(u) du = -\frac{2}{3} \cos(u) + C_1$$ Now substitute 3x back in for u: $$-\frac{2}{3} \cos(3x) + C_1$$

Integrate both sides again

Now that we have f'(x) = -2/3 cos(3x) + C_1, we will integrate both sides again with respect to x: $$\int f'(x) dx = \int -\frac{2}{3} \cos(3x) + C_1 dx$$

Evaluate the integral on the right-hand side

To evaluate the integral on the right-hand side, we will again use integration by substitution. Let v = 3x, so dv/dx = 3 and dx = dv/3: $$\int -\frac{2}{3} \cos(3x) + C_1 dx = -\frac{2}{9} \int \cos(v) dv + C_1 \int 1 dx$$ Now, integrate cos(v) with respect to v and 1 with respect to dx: $$-\frac{2}{9} \int \cos(v) dv + C_1 \int 1 dx = -\frac{2}{9} \sin(v) + C_1x + C_2$$ Now substitute 3x back in for v: $$-\frac{2}{9} \sin(3x) + C_1x + C_2$$

Express the general solution

Finally, we express the general solution as required, f(x) = -2/9 sin(3x) + C_1x + C_2. The general solution is: $$f(x) = -\frac{2}{9} \sin(3x) + C_1x + C_2$$

Key concepts

Integration by Substitution

When faced with complex integrals, integration by substitution is a valuable technique that simplifies the task. It is comparable to the 'change of variables' in algebra. It often involves replacing a portion of the integral with a single variable, simplifying the integral, then reversing the substitution after the integration.

Consider the integral \(\int 2 \sin(3x) dx\). The equation suggests a substitution \(u = 3x\) could simplify it. We calculate derivatives with respect to \(x\) which gives us \(\frac{du}{dx} = 3\) or \(dx = \frac{du}{3}\). Now our integral becomes \(\frac{2}{3} \int \sin(u) du\), an easier task. After integrating, we revert the substitution, replacing \(u\) with \(3x\) to find the function in terms of the original variable \(x\).

Indefinite Integral

An indefinite integral, written as \(\int f(x) dx\), represents a family of functions rather than a single function. This family contains all the antiderivatives of \(f(x)\). An important aspect to note is the \(+ C\), where \(C\) stands for the constant of integration. It's called 'indefinite' because it lacks specific limits as opposed to definite integrals.

In our example, after integrating by substitution, we added the constant \(C_1\) to acknowledge all possible antiderivatives. These constants become essential when you integrate twice, as they cumulatively symbolize multiple integration constants, like \(C_1\) and \(C_2\) in our exercise.

Trigonometric Integration

Trigonometric integration involves integrating functions with trigonometric expressions. These integrals frequently appear in calculus, and recognising them is crucial for applying the right techniques. For instance, the integrals of \(\sin(x)\), \(\cos(x)\), and \(\tan(x)\) follow known formulas.

In our exercise's step 2, integrating \(\sin(u)\) yields \( -\cos(u)\) and in step 4, integrating \(\cos(v)\) yields \(\sin(v)\). These are basic trigonometric integrations essential for the solution. It demonstrates how a substitution can turn a complex integral into a trigonometric function integral, which can be easily solved with these standard formulas.

General Solution of Differential Equation

A differential equation relates a function with its derivatives. The general solution to a differential equation includes all possible solutions or functions that satisfy the equation, often expressed with constants (like \(C_1\) and \(C_2\)) to account for all scenarios.

In the given problem, the second order differential equation \(f''(x) = 2 \sin(3x)\) required us to integrate twice to find the function \(f(x)\). The twice-integrated function that we derived, \(f(x) = -\frac{2}{9} \sin(3x) + C_1x + C_2\), represents the general solution of the original second order differential equation because it encompasses all functions \(f(x)\) that when differentiated twice, will return \(2 \sin(3x)\). These concepts are crucial for understanding not just individual equations but entire families of functions that relate to physical phenomena.