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

1 Answer
Oct 21, 2017

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].