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: $$ y(t) = \frac{\sin t}{t^2} $$