Find the general solution of the following differential equation: xy''+y'=x^2

1. Identify the form of the differential equation

xy'' + y' = x^2

2. This is a linear differential equation. To solve it, we use the method of variation of constants.

3. First, solve the homogeneous equation:

xy'' + y' = 0

4. Simplify to a first-order equation by letting

y' = p

5. Then

y'' = p'

6. Substitute

x p' + p = 0

7. This is separable, so

p' = -\frac{p}{x}

8. Integrate

\ln|p| = -\ln|x| + C_1

9. Which means

p = \frac{C_1}{x}

10. Recall

y' = p = \frac{C_1}{x}

11. Integrate again

y = C_1\ln|x| + C_2

12. Now find the particular solution of the non-homogeneous differential equation using the method of undetermined coefficients.

13. Assume a particular solution in the form of

y_p = Ax^n

14. Find the correct power of \(x\), let's use \(y_p = Ax^3\)

15. Compute

y_p' = 3Ax^2

y_p'' = 6Ax

16. Substitute into the original equation:

x(6Ax) + 3Ax^2 = x^2

17. Simplify and solve for \(A\):

6Ax^2 + 3Ax^2 = x^2

9Ax^2 = x^2

A = \frac{1}{9}

18. Therefore,

y_p=\frac{x^3}{9}

19. Combine the general solution

y = y_h + y_p

20. Result:

y=\frac{x^3}{9}+C_1\ln|x|+C_2