How do you find the general solution of the differential equation x^2y” + 2y’ = e^x?

To find the general solution of the differential equation x²y” + 2y’ = e^x, we first identify the type of equation we are dealing with. This is a linear second-order ordinary differential equation. Here’s a systematic approach to solving it:

1. Rewrite the equation: The given differential equation can be expressed as:

   x²y'' + 2y' - e^x = 0

2. Find the complementary solution (Homogeneous part): We first solve the associated homogeneous equation:

   x²y'' + 2y' = 0

   This is a Cauchy-Euler equation, which can be solved by assuming a solution of the form y = x^m. Plugging this into the homogeneous equation leads to:
   m(m-1)x^m + 2mx^{m-1} = 0
   Simplifying this gives:
   m(m-1) + 2m = 0
   Rearranging, we can factor it as:
   m(m + 1) = 0
   Thus, we have two possible solutions: m = 0 and m = -1. Therefore, the complementary solution is:

   y_c = C₁ + C₂/x

3. Find the particular solution: Next, we look for a particular solution to the non-homogeneous equation. Given that the non-homogeneous term is e^x, we can try a particular solution of the form:

   y_p = Ae^x

   where A is a constant to be determined. Computing the derivatives:

   y_p' = Ae^x,  

   y_p'' = Ae^x

   Substituting y_p into the left side of the original equation, we get:
   x^2(Ae^x) + 2(Ae^x) = e^x. Factoring out e^x, we have:
   e^x(Ax^2 + 2A) = e^x
   Cancelling e^x, this leads to:
   Ax^2 + 2A = 1
   To satisfy this for all x, coefficients must match. Therefore, we get:
   A = 0 and 2A = 1
   Solving gives A = 1/2. Thus, our particular solution is:

   y_p = (1/2)e^x

4. Combine solutions: The general solution is the sum of the complementary solution and the particular solution:

   y = y_c + y_p

   y = C₁ + C₂/x + (1/2)e^x

y = C₁ + C₂/x + (1/2)e^x