What is the general solution of the differential equation x^2 (d^2y)/(dx^2) - 3x dy/dx+y=sin(logx)/x ?
1 Answer
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 = 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,
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
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
e^(-t)(2bsint -2acost) - 4e^(-t)( (a-b)cost - (a+b)sint ) + e^(-t)(asint + bcost) = e^(-t)sint
Equating coefficients of
cos(x): -2a - 4(a-b) + b = 0
sin(x): 2b +4 (a+b) + a = 1
Solving simultaneously we get:
a=5/61 andb=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].