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

Answer 1

see Below

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.

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Answer 2

# y=A/x + Bx^3#

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

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Answer 3

The general solution of the given differential equation (x^2y'' - xy' - 3y = 0) is (y = c_1x^3 + c_2x^{-1}), where (c_1) and (c_2) are arbitrary constants.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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