Question

Find a solution of the differential equation: \(\frac{d y}{d x}=x \cos \left(x^{2}\right) ; y(0)=\pi\)

Short answer

Answer: \(y = \frac{1}{2}\sin\left(x^2\right) + \pi\)

Step-by-step solution

Integrate both sides

Integrate both sides of the equation with respect to \(x\): $$\int \frac{dy}{dx} dx = \int x\cos\left(x^2\right) dx$$ The integral on the left side is just \(y\). So, we have: $$y = \int x\cos\left(x^2\right) dx$$

Integrate the right side using substitution

To integrate the right side, we can use substitution. Let \(u = x^2\), so \(\frac{du}{dx} = 2x\) or \(\frac{1}{2} du = x dx\). Our integral becomes: $$\int x\cos\left(x^2\right) dx = \int \frac{1}{2}\cos(u) du$$ Now, integrate with respect to \(u\): $$\int \frac{1}{2}\cos(u) du = \frac{1}{2}\int \cos(u) du = \frac{1}{2}\sin(u) + C$$ Remembering that \(u = x^2\), we substitute back: $$y = \frac{1}{2}\sin\left(x^2\right) + C$$

Apply the initial condition and solve for C

Use the initial condition \(y(0) = \pi\) to find the constant C. Plug in \(x=0\) and \(y=\pi\): $$\pi = \frac{1}{2}\sin\left(0^2\right) + C$$ We know that \(\sin(0) = 0\), so the equation simplifies to: $$\pi = C$$

Write the final solution

Now that we have found the constant C, we can write the final solution for the given differential equation: $$y = \frac{1}{2}\sin\left(x^2\right) + \pi$$ This is the solution to the differential equation \(\frac{dy}{dx}= x\cos\left(x^2\right)\) with the initial condition \(y(0)= \pi\).

Key concepts

Integration Techniques

When tackling differential equations, integration plays a crucial role. It's the process of finding a function whose derivative matches a given function. In this exercise, we needed to integrate both sides of the equation to find the solution function, y.
On the left side of the differential equation, integrating \(\frac{dy}{dx}\) with respect to \(x\) simply returns \(y\). This is because the integral effectively "undoes" the derivative.
On the right side, however, we face a more complex expression. Integration techniques like substitution, integration by parts, and trigonometric identities can help simplify the process. It's important to choose the right method based on the structure of the expression. Here, substitution was used, as we'll discuss more in the next section.

Substitution Method

The substitution method is a powerful integration technique, especially when dealing with composite functions. It allows the transformation of a difficult integral into a simpler one. In our exercise, the substitution method was key to solving the integral involving trigonometric functions.
We began by choosing \(u = x^2\), which converts our expression into something more manageable. Why \(u = x^2\)? It simplified the cosine expression and matched with \(x\) in the integrand, since \(\frac{du}{dx} = 2x\).
Substitution often involves:

• Identifying a part of the integrand to replace with a single variable like \(u\).
• Determining \(du\) in terms of \(dx\) to replace differentials accordingly.
• Rewriting the entire integral in terms of \(u\).

Once you've rewritten the integral in terms of \(u\), it’s typically much easier to evaluate.

Initial Value Problems

Initial Value Problems (IVPs) are a type of differential equation where a specific condition is given at the start. They’re quite common in physical sciences, where initial conditions model real-life setups, like population sizes or starting velocities.
To solve an IVP, you generally follow these steps:

• First, solve the differential equation as if it were a regular one, ignoring the initial condition.
• Integrate the equation to find the general solution, which often includes an unknown constant (e.g., \(C\)).
• Plug in the initial conditions (such as \(y(0)=\pi\) in our example) to find the specific value for this constant.

In our exercise, we found \(C\) by substituting \(x=0\) and \(y=\pi\) into the solution. Solving \(\pi = \frac{1}{2}\sin(0) + C\) led us to determine that \(C = \pi\). This step is crucial in tailoring the general solution to meet the unique starting criteria of the problem.

Trigonometric Integrals

Trigonometric integrals are a common component of many calculus problems, especially those involving periodic functions like sine and cosine. In the example, our integral included \(\cos(x^2)\), which required careful handling due to the \(x^2\) inside the cosine function.
The integral \(\int \frac{1}{2}\cos(u) du\)\was simplified through substitution. On integration, the cosine function transformed into a sine function, leading to \(\frac{1}{2}\sin(u) + C\).\Finally, by replacing the variable \(u\) with \(x^2\), we returned to the original context of the problem, yielding the integral result as part of our solution. Trigonometric functions often pose a challenge, but the consistent application of substitution or other techniques makes them more manageable.