Problem 20 - Calculus Early Transcendentals

Question

Solve the following problems. $$\frac{d y}{d x}=3 \cos 2 x+2 \sin 3 x, y(\pi / 2)=8$$

Step-by-Step Solution

Answer

To find the original function y(x) that has the derivative $$\frac{d y}{d x}=3 \cos 2 x+2 \sin 3 x$$ and the initial condition $$y(\pi / 2)=8$$, integrate the derivative function. The integrated function is $$y(x) = \frac{3}{2}\sin 2x -\frac{2}{3} \cos 3x + C$$. Applying the initial condition and solving for C, we find that $$C = \frac{26}{3}$$. The final function is $$y(x) = \frac{3}{2}\sin 2x -\frac{2}{3} \cos 3x + \frac{26}{3}$$.

Step 1: Integrate the derivative function

To find the original function, we'll integrate the given derivative function with respect to x: $$ y(x) = \int \left(3 \cos 2 x+2 \sin 3 x\right) dx $$

Step 2: Evaluate the integral

To evaluate the integral, we need to integrate both terms in the integrand separately. Using integration by substitution, we find the following: For the first term, let \(u=2x\), so \(du = 2dx\). Thus, the integral becomes: $$ \frac{3}{2}\int \cos u \, du = \frac{3}{2}\sin u + C_1 = \frac{3}{2}\sin 2x + C_1 $$ For the second term, let \(v=3x\), so \(dv = 3dx\). The integral becomes: $$ \frac{2}{3}\int \sin v \, dv = -\frac{2}{3} \cos v + C_2 = -\frac{2}{3} \cos 3x + C_2 $$ Combining both results, we get the integrated function: $$ y(x) = \frac{3}{2}\sin 2x -\frac{2}{3} \cos 3x + C $$ where \(C = C_1 + C_2\).

Step 3: Apply the initial condition

Now, we'll use the provided initial condition, \(y(\pi / 2)=8\). Plug \(\pi / 2\) into our integrated function: $$ 8 = \frac{3}{2}\sin (2\cdot \frac{\pi}{2}) - \frac{2}{3} \cos (3\cdot \frac{\pi}{2}) + C $$

Step 4: Solve for the constant C

Evaluate the trigonometric functions and simplify the equation: $$ 8 = \frac{3}{2}\sin \pi - \frac{2}{3} \cos \frac{3\pi}{2} + C $$ $$ 8 = 0 - \frac{2}{3}(1) + C $$ After solving for C, we obtain: $$ C = \frac{26}{3} $$

Step 5: Write the final function

Now, plug the constant C back into the integrated function to get the final function: $$ y(x) = \frac{3}{2}\sin 2x -\frac{2}{3} \cos 3x + \frac{26}{3} $$ This is the original function y(x) that the derivative represents, and it satisfies the given initial condition.

Key Concepts

Integration Techniques

Integration is a core technique in calculus. When dealing with differential equations, like the one in our exercise, integrating an expression lets us find the original function given its derivative. Here, the derivative function combines terms like \(3 \cos 2x\) and \(2 \sin 3x\). Each requires careful handling to solve effectively.To solve our given equation, we used a technique known as "u-substitution". This method is crucial for integrating functions that involve composite arguments such as \(2x\) and \(3x\). The main idea is:

• Selecting an inner function, like \(u = 2x\).
• Finding its differential, \(du\), and substituting variables to simplify the integral.

This transforms the integral into a more straightforward form, \(\int \cos u \, du\), which we easily solved to get \(\frac{3}{2} \sin 2x\). The same concept applied to the second term, \(\int \sin v \, dv\), resulting in \(-\frac{2}{3} \cos 3x\). When combined, these results showed a simple approach through substitution integration.

Initial Value Problems

An initial value problem is a type of differential equation where the solution must satisfy a specific condition called an initial value. This condition is given at a particular point, generally at the initial stage of the problem.In the original problem, we knew that the function \(y(x)\) must equal 8 when \(x = \frac{\pi}{2}\). This information is vital as it allows us to determine the constant of integration, \(C\), after completing the integral. Without this step, our solution could have numerous possibilities, since any different value of \(C\) would potentially satisfy the derivative, but not all of them would meet the initial condition.To apply this, we replaced \(x\) with \(\frac{\pi}{2}\) in our solution for \(y(x)\) and solved the equation \(8 = 0 - \frac{2}{3}(1) + C\). Solving this resulted in finding \(C = \frac{26}{3}\), which was then plugged back into our equation to form the full solution satisfying the initial condition.

Trigonometric Integrals

Trigonometric integrals involve the integration of functions with trigonometric expressions, like sinusoidal waves. These are common in calculus due to their periodic nature and vast applications.In our exercise, we encountered an integral with terms \(\cos 2x\) and \(\sin 3x\). Each trigonometric function in an integral requires special considerations depending on their angle multiples; thus, understanding basic integrals of \(\sin \theta\) and \(\cos \theta\) is key:

• \(\int \cos \theta \, d\theta = \sin \theta + C\)
• \(\int \sin \theta \, d\theta = -\cos \theta + C\)

Adjusting these for multiples, like \(2x\) or \(3x\), uses the substitution method discussed earlier, ensuring the integral respects the trigonometric relations correctly. Recognizing the potential to use identities, such as double or triple angle formulas, can simplify future complexities in similar integrals.