What is the general solution of the differential equation x^2y'' -xy'-3y=0 ?

2 Answers
Jun 10, 2017

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 y_1, the second one is given as y_2=v(x)y_1. Where in this problem, y_1=1/x

If y_2=v(x)1/x, then the derivatives are:
y'_2=v(-1/(x^2))+v'(1/x)

and

y''_2=v(2/x^3)+v'(-2/x^2)+v''(1/x)

Substitute this values in the original equation:

x^2y''-xy'-3y=0

And:

x^2(v(2/x^3)+v'(-2/x^2)+v''(1/x))-x(v(-1/(x^2))+v'(1/x))-3(v(1/x))=0

Although this is suprisingly long, this actually simplifies to:

v''-3/xv'=0

Substitute w=v' so that is easy to solve as first order differential equation.

w'-3/xw=0

The integrating factor is 1/x^3(Given by e^(int3/xdx)).

1/x^3w'-3/x^4w=0
(1/x^3w)'=0

Integrate both sides:

1/x^3w=C_1

Undo substitution:

v'1/x^3=C_1
v'=C_1x^3

Integrate both sides again:

v=C_1/4x^4+C_2

Now the general solution is given as:
y_2=v(x)y_1
y_2=(C_1/4x^4+C_2)(1/x)
y_2=(C_1(x^3/4)+C_2(1/x))

Now the first solution is given when C_1=0, then the second solution can be when C_2=0 which is x^3/4 otherwise there are infinite solutionsby just filling the constants.

To check:
y=x^3/4
y'=3/4x^2
y''=3/2x

Substitute:
x^2(3/2x)-x(3/4x^2)-3(x^3/4)=0
3/2x^3-3/4x^3-3/4x^3=0
6x^3-3x^3-3x^3=0
0=0

Which is correct.

Jun 17, 2017

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