5) Solve the equation t²y" + 4ty' +2y=0, knowing that this is an Euler equation.

To solve the given differential equation t²y" + 4ty' + 2y = 0, we can assume a solution of the form y(t) = t^m, where m is a constant. This will allow us to convert the equation into an algebraic equation in terms of m. This is a standard technique for solving Euler equations. First, let's find the first and second derivatives of y(t): y(t) = t^m y'(t) = mt^(m-1) y''(t) = m(m-1)t^(m-2) Now, substitute these derivatives into the original equation: t²y'' + 4ty' + 2y = 0 t²[m(m-1)t^(m-2)] + 4t[mt^(m-1)] + 2t^m = 0 Now, simplify and factor out t^(m-2) (the lowest power of t): m(m-1)t^m + 4mt^m + 2t^m = 0 Next, combine the terms with t^m: t^m [m(m-1) + 4m + 2] = 0 Now, we have an equation involving m: m(m-1) + 4m + 2 = 0 Let's simplify this equation: m^2 - m + 4m + 2 = 0 m^2 + 3m + 2 = 0 Now, we can solve this quadratic equation for m. We can factor it: (m + 2)(m + 1) = 0 So, we have two possible values for m: 1. m + 2 = 0 m = -2 2. m + 1 = 0 m = -1 Now that we have found the values of m, we can write down the two linearly independent solutions for y(t) based on these values: 1. For m = -2: y₁(t) = t^(-2) 2. For m = -1: y₂(t) = t^(-1) So, the general solution to the Euler equation t²y" + 4ty' + 2y = 0 is the linear combination of these two solutions: y(t) = C₁t^(-2) + C₂t^(-1) Where C₁ and C₂ are arbitrary constants that can be determined based on initial conditions or boundary conditions if they are given.