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): $$ y(t) = 3\left(\frac{1}{2}\ln (1 + t^2)\right) - 5(\arctan (t)) + 2 + 5\pi/4 - \frac{3}{2}\ln(2) $$