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Consider the linear, variable-coefficient ODE z2y" 3zy' + 4y = 0, z > 0. (1) 1. Find an integer n such that y;(z) = z is a solution of (1). 2. Look for a second solution ys(z) by setting: y2(z) = y1(z)v(z) = z"v(2), (2) where n is the integer you found in part 1. Show that the original second-order equation (1) reduces to a simpler ODE that can be viewed as a first-order linear equation in v'(z). Solve this equation for v(x), then integrate to determine v(x), and then finally determine the second solution yo(z) using (2). Note #1: You can drop any constants of integration here since you are only really concerned with the functional form of ya(z). Note #2: This trick for using one solution to construct a second is called the \"method of reduction of order\". We will use this method several times throughout the course. 3. Show that {y;(z),y2(x)} form a fundamental solution set. Write the general solution for the ODE in (1). 4. Solve the IVP using the initial conditions y(2)=-2, y'(2)=-1 5. Plot your solution from part 4 on the interval 0