Please need an answer asap

The differential equation thy" - t (t + 2 ) y + (t+ 2 )y=0 has y1 = t as a solution. Applying reduction of order we set y2 = v - y1 - v . t. Then (using the prime notation for the derivatives) y2 = So, substituting y2 and its derivatives into the left side of the differential equation, and reducing, we get tay3 - t(t + 2) 3/2 + (t+ 2)32 = The reduced form has a common factor of " which we can divide out of the equation. Since this equation does not have any v terms in it we can make the substitution u = v giving us the first order linear equation in u: = 0. If we use c as the constant of integration, the solution to this equation is u = Integrating to get v, and then finding y2 gives the general solution: C191 + czy2 = The differential equation ac 2- dy da2 dy + 16y = 0 has a* as a solution. Applying reduction of order we set y2 = ux*. Then (using the prime notation for the derivatives) 32 = yz = So, plugging y2 into the left side of the differential equation, and reducing, we get x y2 - 7xy2 + 1692 = The reduced form has a common factor of a which we can divide out of the equation so that we have ru" + u' = 0. Since this equation does not have any u terms in it we can make the substitution w = u' giving us the first order linear equation aw + w = 0. This equation has integrating factor for x > 0. If we use a as the constant of integration, the solution to this equation is w - Integrating to get u, and using b as our second constant of integration we have u = Finally y2 - and the general solution is