Question

find the solution of the given initial value problem. $$ t y^{\prime}+2 y=\sin t, \quad y(\pi / 2)=1 $$

Short answer

Question: Solve the given initial value problem (IVP) involving a first-order linear ordinary differential equation (ODE) and an initial condition: \(ty'(t) + 2y(t) = \sin(t)\) and \(y(\pi/2) = 1\). Answer: The particular solution for the given initial value problem is: $$ y(t) = \frac{-t^2\cos(t) + 2t\sin(t) + 2\cos(t) + (\pi/2)^2 - 2(\pi/2)}{t^2} $$

Step-by-step solution

Solve the given ODE

First, we need to rewrite the given ODE in standard form: $$ ty'(t) + 2y(t) = \sin(t) $$ To solve this ODE, we will use the integrating factor method. In order to do this, we need to calculate the integrating factor, which is: $$ \rho(t) = e^{\int \frac{2}{t} dt} = e^{2 \ln |t|} = t^2 $$ Now, multiply the ODE by the integrating factor: $$ t^2y'(t) + 2t y(t)=t^2 \sin(t) $$ The left side of this equation is the derivative of \((t^2y(t))'\), so we can rewrite it as: $$ \frac{d(t^2y(t))}{dt} = t^2\sin(t) $$ Now, integrate both sides with respect to \(t\): $$ \int \frac{d(t^2y(t))}{dt} dt = \int t^2\sin(t) dt $$ The left side simplifies to: $$ t^2y(t) $$ The right side can be solved using integration by parts: Let \(u=t^2\), \(dv=\sin(t) dt\). So, \(du = 2t dt\) and \(v=-\cos(t)\). Now, apply integration by parts: $$ uv - \int v du = -t^2\cos(t) + \int 2t\cos(t) dt. $$ We need to apply integration by parts again: Let \(u=2t\), \(dv=\cos(t) dt\). So, \(du=2dt\) and \(v=\sin(t)\). Now, apply integration by parts: $$ 2t\sin(t) - 2\int \sin(t) dt = 2t\sin(t) + 2\cos(t) $$ Substitute this back into the previous integration equation: $$ t^2y(t) = -t^2\cos(t) + 2t\sin(t) + 2\cos(t) + C $$ Now, divide by \(t^2\) to find the general solution: $$ y(t) = \frac{-t^2\cos(t) + 2t\sin(t) + 2\cos(t) + C}{t^2} $$

Use the initial condition to find the particular solution

We are given the initial condition \(y(\pi/2) = 1\). Substitute this into the general solution: $$ 1 =\frac{ -(\pi/2)^2\cos(\pi/2) + 2(\pi/2)\sin(\pi/2) + 2\cos(\pi/2) + C}{(\pi/2)^2} $$ Since \(\cos(\pi/2) = 0\) and \(\sin(\pi/2) = 1\), we simplify the equation to: $$ 1 = \frac{2(\pi/2) + C}{(\pi/2)^2} $$ Solve for \(C\): $$ C = (\pi/2)^2 - 2(\pi/2) $$ Now plug this value of \(C\) into the general solution to obtain the particular solution: $$ y(t) = \frac{-t^2\cos(t) + 2t\sin(t) + 2\cos(t) + (\pi/2)^2 - 2(\pi/2)}{t^2} $$

Key concepts

Ordinary Differential Equations

Ordinary Differential Equations (ODEs) are fundamental in mathematics and applied sciences, as they model various phenomena from physics to economics. An ODE is an equation containing a function of one independent variable and its derivatives. The goal is to find a function that satisfies the given equation.

In our problem, the ODE \( t y'(t) + 2 y(t) = \sin(t) \) involves the time \( t \) as the independent variable and its solution \( y(t) \) as the dependent variable. It's a first-order linear ODE where \( y' \) represents the derivative of \( y \) with respect to \( t \). Solving ODEs often requires using methods tailored to their specific types, such as separation of variables, integrating factor method, or transformation techniques.

Integrating Factor Method

The integrating factor method is a powerful technique for solving linear ordinary differential equations. Its main purpose is to transform the original ODE into an integrable form. The essence of this method lies in finding a function, known as the integrating factor, which, when multiplied with the ODE, allows the left side of the equation to be expressed as the derivative of a product.

In the exercise provided, the integrating factor is \( \rho(t) = e^{\int \frac{2}{t} dt} = t^2 \) which, after being multiplied by the ODE, lets us rewrite it as \( \frac{d(t^2y(t))}{dt} \) on the left side. Notably, this simplification is key to reaching a solvable equation. It’s essential for students to learn this method as many first-order linear ODEs can be efficiently solved using this approach.

Integration by Parts

Integration by parts is a technique based on the product rule for differentiation, which is instrumental in integrating products of functions. It essentially allows us to transform the integral of a product into a set of simpler integrals. The formula for integration by parts is \( \int u dv = uv - \int v du \).

When solving our exercise, integration by parts is used twice to handle the term \( \int t^2\sin(t) dt \). First, we set \( u=t^2 \) and \( dv=\sin(t) dt \), and then again with \( u=2t \) and \( dv=\cos(t) dt \). This step-by-step method reduces the complex integral into simpler forms that we can solve easily. Mastering this technique is crucial for students, as it's widely applicable in calculus for integrating a wide array of functions where other methods fall short.