Finding a Second Linearly Independent Solution Using Reduction of Order

In summary, the given problem asks for the use of reduction of order to find another linearly independent solution to the given differential equation. However, the given function y1 = x-1/2cosx does not satisfy the differential equation as given. The proper form of the equation is x2y" + xy' + (x2 - 1/4)y = 0. The method of reduction of order involves looking for a solution of the form y2 = u(x)x^(1/2)cosx and setting up a first order equation for v = u'.
  • #1
Mark Brewer
38
4

Homework Statement


(Reduction of order) The function y1 = x-1/2cosx is one solution to the differential equation x2y" + xy' + (x2 - 1/4) = 0. Use the method of reduction of order to find another linearly independent solution.

The Attempt at a Solution



I divided x2 to both sides to get the equation into y" + py' + qy = 0

y" + (1/x)y' + ((1 - (1/4x2)) = 0

Using Abel's method

c e-∫(p)dx

= c e-ln x
y2 = -c lnx[/B]

Am I doing this right?
 
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  • #2
Mark Brewer said:

Homework Statement


(Reduction of order) The function y1 = x-1/2cosx is one solution to the differential equation x2y" + xy' + (x2 - 1/4) = 0. Use the method of reduction of order to find another linearly independent solution.

The Attempt at a Solution



I divided x2 to both sides to get the equation into y" + py' + qy = 0

y" + (1/x)y' + ((1 - (1/4x2)) = 0

Using Abel's method

c e-∫(p)dx

= c e-ln x
y2 = -c lnx[/B]

Am I doing this right?
Take y2 and plug it back into the original differential equation and see if it is satisfied. :wink:
 
  • #3
Mark Brewer said:

Homework Statement


(Reduction of order) The function y1 = x-1/2cosx is one solution to the differential equation x2y" + xy' + (x2 - 1/4) = 0. Use the method of reduction of order to find another linearly independent solution.

The Attempt at a Solution



I divided x2 to both sides to get the equation into y" + py' + qy = 0

y" + (1/x)y' + ((1 - (1/4x2)) = 0

Using Abel's method

c e-∫(p)dx

= c e-ln x
y2 = -c lnx[/B]

Am I doing this right?
No, the problem said to use the method of reduction of order.

But first, did you type the differential equation and solution correctly? I ask because the supposed solution doesn't satisfy the given differential equation.
 
  • #4
Mark Brewer said:

Homework Statement


(Reduction of order) The function y1 = x-1/2cosx is one solution to the differential equation x2y" + xy' + (x2 - 1/4) = 0.
You mean x2y" + xy' + (x2 - 1/4)y = 0, don't you? As Vela pointed out, the given y1 does not satisfy the equation you gave. Perhaps it satisfies this one. I haven't checked.

Use the method of reduction of order to find another linearly independent solution.
So you start by looking for a solution of the form [itex]y_2= u(x)x^{1/2}cos(x)[/itex]
Find the first and second derivatives of that and put them into the equation. If your given function really is a solution to the differential equation, then all terms involving only "u" (as opposed to u' or u'') will cancel leaving a first order equation for v= u'.

3. The Attempt at a Solution

I divided x2 to both sides to get the equation into y" + py' + qy = 0

y" + (1/x)y' + ((1 - (1/4x2)) = 0

Using Abel's method

c e-∫(p)dx

= c e-ln x
y2 = -c lnx
Am I doing this right?
Mark Brewer said:

Homework Statement


(Reduction of order) The function y1 = x-1/2cosx is one solution to the differential equation x2y" + xy' + (x2 - 1/4) = 0. Use the method of reduction of order to find another linearly independent solution.

The Attempt at a Solution



I divided x2 to both sides to get the equation into y" + py' + qy = 0

y" + (1/x)y' + ((1 - (1/4x2)) = 0

Using Abel's method

c e-∫(p)dx

= c e-ln x
y2 = -c lnx[/B]

Am I doing this right?
 

What is a second order ODE homogenous equation?

A second order ordinary differential equation (ODE) is an equation that contains a second derivative of an unknown function. A homogenous equation is a type of ODE in which all terms are of the same order and can be expressed in terms of the dependent variable and its derivatives.

What is the general solution to a second order ODE homogenous equation?

The general solution to a second order ODE homogenous equation is a function that satisfies the equation for all possible values of the independent variable. It contains two arbitrary constants, which can be determined by applying initial conditions.

How do you solve a second order ODE homogenous equation?

To solve a second order ODE homogenous equation, you can use the method of undetermined coefficients or the method of variation of parameters. Both methods involve finding a particular solution and adding it to the general solution to get the complete solution.

What is the importance of second order ODE homogenous equations in science?

Second order ODE homogenous equations are important in science because they can model a wide range of physical phenomena, such as oscillations, vibrations, and growth or decay processes. They are also used in many areas of engineering and physics, including mechanics, electricity and magnetism, and quantum mechanics.

How can second order ODE homogenous equations be applied to real-world problems?

Second order ODE homogenous equations can be applied to real-world problems by using mathematical techniques to solve them and obtain solutions that describe the behavior of a system over time. These solutions can then be used to make predictions and analyze the dynamics of the system, which can be helpful in fields such as biology, economics, and environmental science.

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