Question

Find the solution of the following initial value problems. $$h^{\prime}(t)=6 \sin 3 t ; h(\pi / 6)=6$$

Short answer

Question: Find the solution of the initial value problem given the first-order differential equation \(h^{\prime}(t)=6 \sin 3t\) with an initial value \(h(\pi/6)=6\). Answer: The solution of the initial value problem is \(h(t)= -2 \cos(3t) + 6\).

Step-by-step solution

Solving the Differential Equation

We are given the differential equation \(h^{\prime}(t)=6 \sin 3t\). To find the function \(h(t)\), we need to integrate both sides of the equation with respect to \(t\): $$ \int h^{\prime}(t) dt = \int 6 \sin 3t dt. $$

Integrating the Function

On the left side, we have the integral of \(h^{\prime}(t)\), which is simply \(h(t)\). On the right side, we have the integral of \(6 \sin 3t\). The integration for this function requires a \(u\)-substitution: let \(u=3t\), so \(du=3dt\) or \(dt=du/3\): $$ h(t) = \int 6 \sin 3t dt = \int 6 \sin u \frac{du}{3}. $$ Now, we can simplify the integral and find its value: $$ h(t) = 2 \int \sin u du = -2 \cos u + C = -2 \cos(3t) + C. $$

Applying the Initial Value

Now we apply the initial value \(h(\pi/6)=6\) to find the constant \(C\): $$ 6 = -2 \cos(3(\pi/6)) + C. $$ Evaluate \(\cos(3(\pi/6))\): $$ \cos(3(\pi/6)) = \cos(\pi/2) = 0. $$ So, the constant \(C\) is: $$ 6 = -2 \cdot 0 + C \implies C = 6. $$

Writing the Final Solution

Now we have the constant \(C\), and we can write the final solution for the initial value problem: $$ h(t) = -2 \cos(3t) + 6. $$ The solution of the initial value problem is \(h(t)= -2 \cos(3t) + 6\).

Key concepts

Initial Value Problem

An Initial Value Problem (IVP) is a type of differential equation accompanied by specified values, known as initial conditions. These conditions are provided to find specific solutions that satisfy both the differential equation and the initial values. In our problem, we have a function's derivative \( h'(t) = 6 \sin 3t \) and an initial condition \( h(\frac{\pi}{6}) = 6 \).

This means we are tasked with finding an original function \( h(t) \) whose derivative equals the given function, and which also passes through the point \( (\frac{\pi}{6}, 6) \).

To solve such problems, you'll typically go through the following steps:

• Differential Equation: Start with the given differential equation.
• Integration: Integrate the equation to find the original function.
• Application of Initial Conditions: Use given conditions to find any constants introduced during integration.
• Solution: Write the complete solution satisfying both the differential equation and the initial condition.

When an initial value is applied, it essentially selects the unique solution among the possible family of solutions to the differential equation.

Integration by Substitution

Integration by substitution is a method used to simplify complex integrals. It works by transforming the original variable of integration into a new variable, often simplifying the integral into a more manageable form. In our exercise, the integral to solve is \( \int 6 \sin 3t \, dt \).

To make the integration easier, we use a substitution:

• Let \( u = 3t \). This substitution simplifies the trigonometric expression by reducing the multiple of \( t \).
• Then, differentiate \( u = 3t \) to find \( du = 3 \, dt \), so \( dt = \frac{du}{3} \).

Substituting these into the integral gives us \( \int 6 \sin u \frac{du}{3} \), which simplifies to \( 2 \int \sin u \, du \). This is much easier to integrate.

The integration yields \(-2 \cos u + C\), where \( C \) is the integration constant. Substituting back \( u = 3t \), we find \( h(t) = -2 \cos(3t) + C \). This technique proves invaluable in calculus when dealing with constants or functions within an integral that can be simplified through substitution.

Trigonometric Functions

Trigonometric functions, such as sine and cosine, are fundamental in mathematics, often describing periodic phenomena. In this problem, we deal with the function \( \sin 3t \). The solution process reveals how trigonometric identities and evaluations are used in calculus to solve differential equations.

For the integration process, we used \( \sin \) and \( \cos \), leveraging known integrals:

• The integral of \( \sin(u) \) is \(-\cos(u)\).
• Therefore, \( 2 \int \sin u \, du = -2 \cos u \).

Subsequently, during the application of the initial condition \( h(\pi/6) = 6 \), it was vital to evaluate the cosine function.

In the solution, we calculated \( \cos(\pi/2) = 0 \). This evaluation determined the value of the constant \( C \), which completes the specific solution given the initial condition. Understanding these trigonometric properties ensures that you can manipulate and solve differential equations involving periodic functions effectively.