Question

$$ \left(1+t^{2}\right) y^{\prime \prime}+2 t y^{\prime}+3 t^{-2}=0, \quad y(1)=2, \quad y^{\prime}(1)=-1 $$

Short answer

Based on the step by step solution provided, find the function y(t) that satisfies the given second-order linear homogeneous differential equation with the initial conditions: The function y(t) that satisfies the given differential equation and its initial conditions is: $$ y(t) = 3\left(\frac{1}{2}\ln (1 + t^2)\right) - 5(\arctan (t)) + 2 + 5\pi/4 - \frac{3}{2}\ln(2) $$

Step-by-step solution

Rewrite the given equation

Rewrite the given second-order linear homogeneous differential equation in terms of y(t), y'(t), and y''(t): $$ (1 + t^2)y''(t) + 2ty'(t) + 3t^{-2} = 0 $$

Solve for the Integrating Factor

In this step, we'll find the integrating factor by solving the differential equation for the integrating factor (IF): $$ \mu(t) = e^{\int \frac{2t}{1+t^2} dt} $$ To integrate the expression inside the exponent, use substitution. Let \(u = 1 + t^2\), so \(du = 2t dt\). Thus, the integrating factor becomes: $$ \mu(t) = e^{\int \frac{du}{u}} = e^{\ln |u| + C}= u = 1 + t^2 $$ Now, multiply the original equation by the integrating factor: $$ (1 + t^2)^2y''(t) + 2t(1 + t^2)y'(t) + 3 = 0 $$

Integrate the Resulting Equation

Now, notice that the left-hand side of the equation is an exact derivative: $$ \frac{d}{dt} \left[(1+t^2)y'(t)\right] = (1 + t^2)^2y''(t) + 2t(1 + t^2)y'(t) $$ Therefore, the resulting equation is: $$ \frac{d}{dt} \left[(1+t^2)y'(t)\right] - 3 = 0. $$ Integrating both sides with respect to t, we have: $$ (1+t^2)y'(t) = 3t + C_1 $$

Apply Initial Conditions and Find Constants

Now, apply the initial conditions to find the constants involved: $$ y'(1) = -1, y(1) = 2 $$ From the expression for y'(t), we get: $$ (1+1)(-1) = 3(1) + C_1 \Rightarrow C_1 = -5 $$ Thus, the equation for y'(t) becomes: $$ y'(t) = \frac{3t - 5}{1 + t^2} $$ Now, integrate y'(t) with respect to t to obtain y(t): $$ y(t) = \int \frac{3t - 5}{1 + t^2} dt $$ Separate the integral into two parts: $$ y(t) = 3\int \frac{t}{1 + t^2} dt - 5\int \frac{1}{1 + t^2} dt $$ Using substitution and the arctangent formula: $$ y(t) = 3\left(\frac{1}{2}\ln (1 + t^2)\right) - 5(\arctan (t)) + C_2 $$ Finally, apply the initial condition y(1) = 2: $$ 2 = 3\left(\frac{1}{2}\ln (2)\right) - 5(\arctan (1)) + C_2 \Rightarrow C_2 = 2 + 5\pi/4 - \frac{3}{2}\ln(2) $$

Write the Final Solution

Now, we have found all the constants and can write the final solution y(t): $$ y(t) = 3\left(\frac{1}{2}\ln (1 + t^2)\right) - 5(\arctan (t)) + 2 + 5\pi/4 - \frac{3}{2}\ln(2) $$

Key concepts

Integrating Factor

When solving linear non-homogeneous differential equations, the integrating factor (IF) is a useful method. It transforms the equation into a form where we can easily integrate both sides. Here, we aim to solve a differential equation of the form:\[ ay'' + by' + cy = f(t) \]To find the integrating factor \( \mu(t) \), we use:- The formula \( \mu(t) = e^{\int P(t) \, dt} \), where \( P(t) \) is the coefficient of \( y' \) divided by the coefficient of \( y \) in the equation.- In our case, the equation rearranges to fit this form, and using substitution helps find \( \mu(t) = 1 + t^2 \).Thus, multiplying the entire equation by this integrating factor simplifies it into an exact derivative on the left-hand side.

Initial Conditions

Initial conditions are specific values provided for a differential equation that allow us to find particular solutions. In the provided differential equation:- We have \( y(1) = 2 \) and \( y'(1) = -1 \).These initial conditions are crucial because they help us determine the constant(s) of integration when solving the differential equation. They ensure that our solution will not just be a general solution, but rather the unique solution that also satisfies the specific 'real-world' conditions given in the problem.By applying these initial conditions to our solutions for \( y'(t) \) and \( y(t) \), we can calculate the constants \( C_1 \) and \( C_2 \) accurately.

Exact Derivative

The term 'exact derivative' refers to an expression that is the derivative of some function. In solving differential equations, identifying an exact derivative is a powerful technique as it simplifies the problem significantly.When the differential equation is multiplied by the integrating factor, it becomes possible to express the left-hand side as:\[ \frac{d}{dt} \left[(1 + t^2)y'(t)\right] \]Recognizing this expression as an exact derivative allows us to integrate both sides easily. This leads directly to solving the differential equation.The process streamlines solving steps since direct integration of exact derivatives results in simpler expressions to work with.

Arctangent Formula

The arctangent formula emerges when integrating functions of the form \( \frac{1}{1+t^2} \). It is an essential integral:- The integral \( \int \frac{1}{1+t^2} \, dt = \arctan(t) + C \).This formula is particularly useful in solving differential equations involving such terms, as it appears naturally in many contexts, especially those involving trigonometric relationships.During our solving process, we encounter \( \int \frac{1}{1+t^2} \, dt \), requiring the arctangent formula to progress from the expression for \( y'(t) \) to \( y(t) \).It helps simplify the sequence of steps needed to reach a closed-form particular solution that adheres to the initial conditions given in the differential equation problem.