Question

find the solution of the given initial value problem. $$ y^{\prime}+(2 / t) y=(\cos t) / t^{2} \quad y(\pi)=0 $$

Short answer

Question: Solve the initial value problem (IVP) \(y^{\prime}+(2 / t) y=(\cos t) / t^{2}\), with \(y(\pi)=0\). Answer: The particular solution for the given IVP is \(y(t) = \dfrac{\sin t}{t^2}\).

Step-by-step solution

Identify the equation form and the functions p(t) and q(t)

The given differential equation is: $$ y^{\prime}+(2 / t) y=(\cos t) / t^{2} $$ In this case, we see that \(p(t) = \frac{2}{t}\) and \(q(t) = \frac{\cos t}{t^2}\).

Calculate the integrating factor

To find the integrating factor, \(\mu(t)\), we use the formula: $$ \mu(t) = e^{\int p(t) dt} $$ Substitute \(p(t) = \frac{2}{t}\) and find the integral: $$ \mu(t) = e^{\int \frac{2}{t} dt} = e^{2\ln|t|} = t^2 $$

Multiply the integrating factor by the original equation

Now, multiply the integrating factor \(\mu(t) = t^2\) to the original ODE: $$ t^2 y^{\prime} + 2t y = \cos t $$

Rewrite the left-hand side as a derivative

The left-hand side of the modified ODE can be written as the derivative of the product of the integrating factor and the dependent variable (y). Therefore, we can rewrite the equation as: $$ \frac{d}{dt}(t^2y) = \cos t $$

Integrate both sides of the equation

Integrate both sides of the modified ODE with respect to t: $$ \int \frac{d}{dt}(t^2y) dt = \int \cos t \, dt \\ t^2y = \sin t + C $$

Solve for y

Now, we can solve for y by dividing both sides by the integrating factor: $$ y(t) = \frac{\sin t + C}{t^2} $$

Apply the initial condition

Lastly, apply the initial condition \(y(\pi) = 0\) to find the particular solution: $$ y(\pi) = \frac{\sin \pi + C}{\pi^2} = 0 $$ This results in \(C=0\), as \(\sin \pi=0\). Thus, the particular solution is: $$ y(t) = \frac{\sin t}{t^2} $$

Key concepts

Differential Equation

A differential equation is an equation that relates a function to its derivatives. In this exercise, we're looking at a first-order linear differential equation of the form \( y' + p(t)y = q(t) \).
Here, the functions \( p(t) \) and \( q(t) \) are known, and we need to find the function \( y(t) \) that satisfies the equation. This type of equation is common in modeling physical processes.
When approaching such problems, it's crucial to recognize the equation structure and identify the components \( p(t) \) and \( q(t) \), which we've done as \( p(t) = \frac{2}{t} \) and \( q(t) = \frac{\cos t}{t^2} \).
This identification helps in finding the integrating factor, which is a vital step in solving the equation.

Integrating Factor

The integrating factor is a mathematical technique used to simplify the solving of a linear first-order differential equation.
It transforms the equation into a form that is easier to integrate. To find it, we use the formula:

• Integrating Factor: \( \mu(t) = e^{\int p(t) dt} \)

For our equation, substituting \( p(t) = \frac{2}{t} \) gives:
\( \mu(t) = e^{\int \frac{2}{t} dt} = e^{2\ln|t|} = t^2 \).
Multiplying the entire original equation by this integrating factor, \( t^2 \), significantly simplifies the process of finding the solution. The method essentially converts the left-hand side into the derivative of a product, which is a more straightforward expression to handle.

Particular Solution

In the general solution of a differential equation, we often include an arbitrary constant. However, a particular solution is specific to the problem at hand and is often determined by a given initial condition.
The particular solution satisfies both the differential equation and the initial condition. In this exercise, after transforming the equation using an integrating factor and integrating both sides, we arrived at:
\( t^2y = \sin t + C \).
Solving for \( y(t) \), we get:
\( y(t) = \frac{\sin t + C}{t^2} \).
To find the particular solution, we use the given initial condition which allows us to solve for the constant \( C \), thus providing a solution specific to the conditions of the problem.

Initial Condition

An initial condition in the context of differential equations is a specific point at which the values of the function and its derivatives are known. This is used to find the particular solution of the differential equation.
Here, the initial condition is given as \( y(\pi) = 0 \). To apply it, we substitute \( t = \pi \) into our expression for \( y(t) = \frac{\sin t + C}{t^2} \).
The aim is to determine the constant \( C \).
Substituting, we have \( \frac{\sin \pi + C}{\pi^2} = 0 \), which simplifies to \( C = 0 \) since \( \sin \pi = 0 \).
Thus, the equation resolves to \( y(t) = \frac{\sin t}{t^2} \), ensuring the solution is specific and correct for the given initial conditions.