What is the general solution of the differential equation x^2y'' -xy'-3y=0 ?
2 Answers
see Below
Explanation:
There is a certain way to find the general solution if a one solution is already given. It is said that if given solution
If
and
Substitute this values in the original equation:
And:
Although this is suprisingly long, this actually simplifies to:
Substitute
The integrating factor is
Integrate both sides:
Undo substitution:
Integrate both sides again:
Now the general solution is given as:
Now the first solution is given when
To check:
Substitute:
Which is correct.
y=A/x + Bx^3
Explanation:
We have:
x^2y'' -xy'-3y=0 ..... [A]
This is a Euler-Cauchy Equation 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:
x^2((d^2y)/(dt^2)-dy/dt)e^(-2t) -xe^(-t)dy/dt-3y=0
:. ((d^2y)/(dt^2)-dy/dt) -dy/dt-3y=0
:. (d^2y)/(dt^2)-2dy/dt-3y=0 ..... [B]
This is now a second order linear homogeneous Differentiation Equation. The standard approach is to look at the Auxiliary Equation, which is the quadratic equation with the coefficients of the derivatives, i.e.
m^2-2m-3 = 0
We can solve this quadratic equation, and we get two real and distinct solutions:
(m+1)(m-3) = 0 => m=-1,3
Thus the Homogeneous equation [B]:
:. (d^2y)/(dt^2)-2dy/dt-3y=0
has the solution:
y=Ae^(-t) + Be^(3t)
Now we initially used a change of variable:
x = e^t => t=lnx
So restoring this change of variable we get:
y=Ae^(-lnx) + Be^(3lnx)
:. y=Ae^(lnx^(-1)) + Be^(lnx^3)
:. y=Ax^(-1) + Bx^3
:. y=A/x + Bx^3
Which is the General Solution