Reduction of order in solving second order differential equations

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chwala
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TL;DR Summary
Why is the constant dropped when determining solutions to second order differential equations. (See highlight in red -attached). Otherwise, the reduction of order approach is pretty straightforward.
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chwala said:
TL;DR Summary: Why is the constant dropped when determining solutions to second order differential equations. (See highlight in red -attached). Otherwise, the reduction of order approach is pretty straightforward.

View attachment 338047
Because keeping ##k## is the same thing as adjusting the value of the arbitrary constant ##c_1## in the general solution?
 
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renormalize said:
Because keeping ##k## is the same thing as adjusting the value of the arbitrary constant ##c_1## in the general solution?
Thanks noted...was wondering why they substituted for the constant ##c=-3## and dropped the other constant ##k##. The constant ##c## was not dropped as indicated rather a value was assigned to it. .

Is this not for convenience? to perhaps have " nice solutions'.

Cheers man.
 
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chwala said:
Thanks noted...was wondering why they substituted for the constants ##c=-3## and dropped ##k##. The constant ##c## was not dropped as indicated rather a value was assigned to it. .
Just like keeping ##k## amounts to redefining ##c_1##, wouldn't keeping ##-c/3## simply adjust the value of the arbitrary constant ##c_2##? These are the types of simplifications that come naturally once you're practiced enough solving differential questions. When solving a 2nd-order ODE, as long as you're left in the end with two arbitrary constants, you know you've found the general solution.
 
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1. How does reduction of order work in solving second order differential equations?

Reduction of order involves assuming a solution of the form y = v(x)u(x), where v(x) is a known function and u(x) is an unknown function to be determined. By substituting y = v(x)u(x) into the second order differential equation, we can reduce it to a first order differential equation in terms of u(x).

2. When is reduction of order used in solving second order differential equations?

Reduction of order is typically used when one solution of the homogeneous second order differential equation is known. By assuming a second linearly independent solution in the form of y = v(x)u(x), we can reduce the order of the differential equation and find the second solution.

3. What are the steps involved in reduction of order for second order differential equations?

The steps involved in reduction of order include assuming a solution of the form y = v(x)u(x), substituting it into the differential equation, solving for u(x) by reducing the order of the differential equation, and finding the general solution by combining the known solution with the newly found solution.

4. Can reduction of order be used for nonhomogeneous second order differential equations?

Reduction of order is typically used for homogeneous second order differential equations, where the right-hand side of the equation is zero. For nonhomogeneous equations, other methods such as variation of parameters or undetermined coefficients are more appropriate.

5. What are the advantages of using reduction of order in solving second order differential equations?

Reduction of order can simplify the process of solving second order differential equations by reducing them to first order differential equations. This method is particularly useful when one solution is known, as it allows us to find a second linearly independent solution easily.

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