Question

What is the general solution of the differential equation? : `Acosx + Bsinx -sinxln|cscx+cotx|`

Answer 1

`Acosx + Bsinx -sinxln|cscx+cotx|`

Explanation:

We have:

`y'' + y = cot(x)` ..... [A]

This is a second order linear non-Homogeneous Differentiation Equation. The standard approach is to find a solution, `y_c` of the homogeneous equation by looking at the Auxiliary Equation, which is the quadratic equation with the coefficients of the derivatives, and then finding an independent particular solution, `y_p` of the non-homogeneous equation.

Complementary Function

The homogeneous equation associated with [A] is

`y'' + 9y = 0`

And it's associated Auxiliary equation is:

`m^2 + 1 = 0`

Which has pure imaginary solutions `m=+-i`

Thus the solution of the homogeneous equation is:

`y_c = e^0(Acos(1x) + Bsin(1x))`
`\ \ \ = Acosx + Bsinx`

Particular Solution

With this particular equation [A], the interesting part is find the solution of the particular function. We would typically use practice & experience to "guess" the form of the solution but that approach is likely to fail here. Instead we must use the Wronskian. It does, however, involve a lot more work:

Once we have two linearly independent solutions say `y_1(x)` and `y_2(x)` then the particular solution of the general DE;

`ay'' +by' + cy = p(x)`

is given by:

`y_p = v_1y_1 + v_2y_2 \ \ `, which are all functions of `x`

Where:

`v_1 = -int \ (p(x)y_2)/(W[y_1,y_2]) \ dx`
`v_2 = \ \ \ \ \ int \ (p(x)y_1)/(W[y_1,y_2]) \ dx`

And, `W[y_1,y_2]` is the wronskian; defined by the following determinant:

`W[y_1,y_2] = | ( y_1,y_2), (y'_1,y'_2) |`

So for our equation [A]:

`p(x) = cotx`
`y_1 \ \ \ = cosx => y'_1 = -sinx`
`y_2 \ \ \ = sinx => y'_2 = cosx`

So the wronskian for this equation is:

`W[y_1,y_2] = | ( cosx,,sinx), (-sinx,,cosx) |`

`" " = (cosx)(cosx) - (sinx)(-sinx)`
`" " = cos^2 x + sin^2 x`
`" " = 1`

So we form the two particular solution function:

`v_1 = -int \ (p(x)y_2)/(W[y_1,y_2]) \ dx`

`\ \ \ = -int \ (cotx sinx)/1 \ dx`
`\ \ \ = -int \ cosx/sinx sinx \ dx`
`\ \ \ = -int \ cosx \ dx`
`\ \ \ = -sinx`

And;

`v_2 = \ \ \ \ \ int \ (p(x)y_1)/(W[y_1,y_2]) \ dx`

`\ \ \ = int \ (cotx cosx )/1\ dx`
`\ \ \ = int \ cosx/sinx cosx \ dx`
`\ \ \ = int \ cos^2x/sinx \ dx`
`\ \ \ = int \ (1-sin^2x)/sinx \ dx`
`\ \ \ = int \ cscx-sinx \ dx`
`\ \ \ = -ln|cscx+cotx|+cosx`

And so we form the Particular solution:

`y_p = v_1y_1 + v_2y_2`
`\ \ \ = (-sinx)(cosx) + (cosx-ln|cscx+cotx|)sinx`

Which then leads to the GS of [A}

`y(x) = y_c + y_p`
`\ \ \ \ \ \ \ = Acosx + Bsinx -sinxcosx + sinxcosx-sinxln|cscx+cotx|`
`\ \ \ \ \ \ \ = Acosx + Bsinx -sinxln|cscx+cotx|`