# Solve the coordinate equation x* (d^2y/dx^2) + dy/ dx = x^2

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In this problem we can see that there is no use of the dependent variable y in the equation (except in the derivatives)

To solve it, we first reduce the equation to a simpler form by using the substitution p = dy /dx.

p = dy/dx => dp / dx = (d^2y / dx^2)

The equation is now reduced to x* (dp/dx) + p = x^2

Multiply all the terms by dx , we get: x*dp + p*dx = x^2*dx

Now x*dp + p*dx = d(xp) and x^2*dx = d(x^3/3)

So, we have d(xp) = d(x^3/3)

Integrating, we get: xp = x^3/3 + C

Replace the p by dy/ dx

x* (dy/dx) = x^3/3 + C

=> dy/dx = x^2/3 + C/x

Integrating

y = x^3 / 9 + C1*ln|x| + C2

**Therefore the required solution is y = x^3/9 + C1* ln|x| + C2**

You need to consider the derivative `(dy)/(dx)` as a function `g'(x)` and x as the function `f(x)` , such that:

`f(x)*(g'(x))' + f'(x)*g(x) = x^2`

You should notice that the left side summation represents the product rule, that is used when you need to evaluate the derivative of the product of two functions, such that:

`f(x)*(g'(x))' + f'(x)*g(x) = (f(x)*g'(x))'`

Hence, the equation you need to solve is the following, such that:

`(f(x)*g'(x))' = x^2`

You need to integrate both sides with respect to x, such that:

`int (f(x)*g'(x))' dx = int x^2 dx`

`f(x)*g'(x) = x^3/3 + c`

Replacing back x for f(x) yields:

`x*g'(x) = x^3/3 + c`

Dividing both sides by x yields:

`g'(x) = (x^3/3 + c)/x => g'(x) = x^2/3 + c/x`

You need to integrate both sides with respect ro x, such that:

`int g'(x) dx = int x^2/3 dx + int c/x dx`

`g(x) = y(x) = x^3/ 9 + c_1*ln x + c_2`

**Hence, evaluating the general solution to the given second order linear ordinary differential equation, yields **`y(x) = x^3/ 9 + c_1*ln |x| + c_2.`