# How to solve this problem? `x^2 y'' + 2xy' -1 = 0` For a second order differential equation of the form y''=f(x,y'), the substitutions v=y' , v'=y'' lead to first order differential equation of the form v'=f(x,v).

The equation you give is a nonhomogeneous form of the Cauchy-Euler equation (see link below).

`x^2y'' + 2xy' = 1`

Now, recognize that as a nonhomogeneous equation, the general solution with be of the form:

`y(x) = y_h + y_p`

where `y_h` is the homogeneous solution and `y_p` is the...

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The equation you give is a nonhomogeneous form of the Cauchy-Euler equation (see link below).

`x^2y'' + 2xy' = 1`

Now, recognize that as a nonhomogeneous equation, the general solution with be of the form:

`y(x) = y_h + y_p`

where `y_h` is the homogeneous solution and `y_p` is the particular solution.

To find `y_h` we simply set the right side to zero and solve for y:

`x^2y'' + 2xy' = 0`

Looking at the derivative, we can see easily that `y_h` is in polynomial form (this is what I suppose Cauchy and Euler noticed, too). Let's suppose that `y = x^m`. If this were the case, `y' = mx^(m-1)` and `y'' = m(m-1)x^(m-2)`

Substituting these two values into our differential equation:

`x^2(m(m-1)x^(m-2)) + 2x(mx^(m-1)) = 0`

Simplifying:

`m(m-1)x^m + 2mx^m = 0`

We can divide through by `x^m` in order to eventually solve for m:

`m(m-1) + 2m = 0`

`m(m-1+2) = 0`

`m(m+1) = 0`

Therefore, we get two values for m: 0 and -1

Our function `y_h` will be: `y_h = c_1 + c_2x^-1`

Now, to solve for the nonhomogeneous part, we simply use the method of variation of parameters to find our particular solution:

`y_p = -y_1int((y_2)n(x))/Wdx + y_2int((y_1)n(x))/W dx'`

where `y_1` and `y_2` are the bases of your homogeneous solution, W is the Wronskian (see link below) of the two functions, and n(x) is the nonhomogeneous function of x in the D.E.

`y'' + 2/xy' = 1/x^2`

In our case,

`y_1 = c_1`

`y_2 = c_2x^(-1)`

`n(x) = 1/x^2`

Our n(x) is `1/x^2` because in order for variation of parameters to work, we need a coefficient of 1 for y'', so we needed to divide through by `x^2`

Given this info, let's just calculate our Wronskian quickly:

`W = y_1y_2' - y_1'y_2 = c_1(-c_2x^-2) - 0 = -c_1c_2x^-2`

So, let's get our particular solution!

`y_p = -c_1int(c_2x^-1*x^-2)/(-c_1c_2x^-2) dx + c_2x^-1 int (c_1*x^-2)/(-c_1c_2x^-2) dx`

Simplifying:

`y_p = int 1/x dx + x^-1int 1 dx`

`y_p=lnx + K_1 + x^-1* x + K_2x^-1`

`y_p = lnx + 1 + K_1 + K_2x^-1`

Where `K_1, K_2` are our constants of integration. These, don't matter, though, because they just give the homogeneous solution again (including the constant 1 we found), so we can throw away the solution of 1 and let `K_1=K_2=0` with no effect on our solution.

Our overall solution now becomes:

`y = y_h+y_p = c_1 + c_2x^-1 + lnx`

And there is your overall solution. You may use the method you outlined above, too. I'll outline a solution quickly below (using techniques for a nonhomogeneous first order equation).

If we let v = y', our D.E., after we set the coefficient of v' to 1, becomes:

`(dv)/dx + 2/xv=1/x^2`

This has a nice theoretical general solution of the form:

`v(x) = e^(-int(2(dx)/x))[inte^(int(2(dx)/x))*(1/x^2) dx + C]`

Simplifying:

`v(x) = e^(-2lnx)[int e^(2lnx)/x^2 dx +C]`

`v(x) = 1/x^2 [int x^2/x^2 dx + C]`

`v(x) = 1/x^2(x + C) = 1/x + C/x^2`

Recognizing v(x) = y'(x):

`y(x) = int1/x + C/x^2 dx = lnx -C_1/x + C_2`

Recognizing constants of integration can be messed with, we get the same solution we got with the Cauchy-Euler method:

`y(x) = C_1/x + C_2 + lnx`

I hope you got this from one of the two ways I showed!

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