Question

What is the general solution of the differential equation `x^2 (d^2y)/(dx^2) - 3x dy/dx+y=sin(logx)/x`?

Answer 1

`y = Ax^(2-sqrt(3)) + Bx^(2+sqrt(3)) + (5sinlnx)/(61x) + (6coslnx)/(61x)`

Explanation:

We have:

`x^2 (d^2y)/(dx^2) - 3x dy/dx+y=sin(logx)/x` ..... [A]

This is a Euler-Cauchy Equation (the power of `x` is the same as the degree of the differential in every occurrence of their product) which is typically solved via a change of variable. Consider the substitution:

`x = e^t => xe^(-t)=1`

Then we have,

`dy/dx = e^(-t)dy/dt`, and, `(d^2y)/(dx^2)=((d^2y)/(dt^2)-dy/dt)e^(-2t)`

Substituting into the initial DE [A] we get:

`(e^t)^2 ((d^2y)/(dt^2)-dy/dt)e^(-2t) - 3(e^t) e^(-t)dy/dt+y=sin(loge^t)/((e^t))`

`:. (d^2y)/(dt^2)-dy/dt - 3dy/dt+y = e^(-t)sint`

`:. (d^2y)/(dt^2) - 4dy/dt+y = e^(-t)sint` ..... [B]

This is now 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 associated homogeneous equation is:

`(d^2y)/(dt^2) - 4dy/dt+y = 0` ..... [C]

Which has the Auxiliary Equation:

`m^2-4m+1 = 0`

We can solve this quadratic equation, and we get two distinct real solutions::

`m=2+-sqrt(3)`

Thus the Homogeneous equation [C] has the solution:

`y_c = Ae^((2-sqrt(3))t) + Be^((2+sqrt(3))t)`

Particular Solution

With this particular equation [B], a probably solution is of the form:

`y = ae^(-t)sint + be^(-t)cost`
`\ \ = e^(-t)(asint + bcost)`

Where `a` and `b` are constants to be determined. Let us assume the above solution works, in which case be differentiating wrt `x` we have:

`y' \ \= e^(-t)(acost - bsint) - e^(-t)(asint + bcost)`
`\ \ \ \ \= e^(-t)( (a-b)cost - (a+b)sint )`

`y'' = e^(-t)(-(a-b)sint - (a+b)cost ) - e^(-t)((a-b)cost - (a+b)sint )`
`\ \ \ \ \ = e^(-t)(2bsint -2acost)`

Substituting into the Differential Equation `[B]` we get:

`e^(-t)(2bsint -2acost) - 4e^(-t)( (a-b)cost - (a+b)sint ) + e^(-t)(asint + bcost) = e^(-t)sint`

Equating coefficients of `cos(x)` and `sin(x)` we get:

`cos(x): -2a - 4(a-b) + b = 0`
`sin(x): 2b +4 (a+b) + a = 1`

Solving simultaneously we get:

`a=5/61` and `b=6/61`

And so we form the Particular solution:

`y_p = 5/61e^(-t)sint + 6/61e^(-t)cost`

General Solution

Which then leads to the GS of [B}

`y(t) = y_c + y_p`
`\ \ \ \ \ \ \ = Ae^((2-sqrt(3))t) + Be^((2+sqrt(3))t) + 5/61e^(-t)sint + 6/61e^(-t)cost`

Now we initially used a change of variable:

`x = e^t => t=lnx`

So restoring this change of variable we get:

`y = Ae^((2-sqrt(3))lnx) + Be^((2+sqrt(3))lnx) + 5/61e^(-lnx)sinlnx + 6/61e^(-lnx)coslnx`

`:. y = Ax^(2-sqrt(3)) + Bx^(2+sqrt(3)) + 5/61e^(ln(1/x))sinlnx + 6/61e^(ln(1/x))coslnx`

`\ \ \ \ \ \ = Ax^(2-sqrt(3)) + Bx^(2+sqrt(3)) + 5/61 (1/x)sinlnx + 6/61 (1/x)coslnx`

`\ \ \ \ \ \ = Ax^(2-sqrt(3)) + Bx^(2+sqrt(3)) + (5sinlnx)/(61x) + (6coslnx)/(61x)`

Which is the General Solution of [A].