Second Order Differential Nonhomogeneous Equation

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SUMMARY

The forum discussion focuses on solving a second-order differential nonhomogeneous equation using the method of undetermined coefficients. The specific equation involves finding a particular solution, Yp, alongside the complementary solution, Yc, derived from the roots of the characteristic equation, which are -1 and -4. The user attempted to use Yp = (At^2 + Bt + C)e^(4t) but encountered difficulties in determining the constants. Other methods such as the D Operator method and variation of parameters are suggested as alternatives for finding Yp efficiently.

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  • Understanding of second-order differential equations
  • Familiarity with the method of undetermined coefficients
  • Knowledge of complementary and particular solutions
  • Basic skills in calculus, particularly differentiation
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  • Study the variation of parameters method for finding particular solutions
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Students and professionals in mathematics, engineering, and physics who are working on differential equations, particularly those seeking to enhance their problem-solving skills in nonhomogeneous cases.

GogumaDork
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Homework Statement


Use the method of undetermined coefficients to find one solution of
http://img85.imageshack.us/img85/6844/4ab921ad6ba6851cc91401c.png
Note that the method finds a specific solution, not the general one.

Homework Equations


Y = Yc + Yp
Yc = C1e^(r1t)+C2e^(rt) when roots are not the same.

The Attempt at a Solution


For the homogeneous part I used the quadratic equation to get the roots -1 and -4.
Y = Yc + Yp
Yc = C1e^-t +C2e^-4t

I tried Yp = (At^2+Bt+C)*De^(4t) but it doesn't seem correct after solving a bunch of derivatives and plugging it back into y'' + 3y' -4y and solving for constants.

How to determine the Y-particular for this problem?
 
Last edited by a moderator:
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Hii. Method of undetermined coefficients is sometimes lengthy but it sure does give the correct particular integral. As u have already written the rough Yp,what u have to do now is just replace it for y in the original equation and compare the coefficients of (e^4t)t^2,(e^4t)t and e^4t on both sides of the equation and u will get the expression for Yp. There are however other useful and short methods for finding Yp. One is the D Operator method and the other is the variation of parameters method. These are too long for me to describe elaborately here,but u must check these out in the net or any other reference books.Any standard engineering mathematics book would contain these topics.Personally I recommend using the D Operator method as it is very useful when exploited efficiently.
 
And welcome to Physics Forums...
 
GogumaDork said:

Homework Statement


Use the method of undetermined coefficients to find one solution of
http://img85.imageshack.us/img85/6844/4ab921ad6ba6851cc91401c.png
Note that the method finds a specific solution, not the general one.

Homework Equations


Y = Yc + Yp
Yc = C1e^(r1t)+C2e^(rt) when roots are not the same.

The Attempt at a Solution


For the homogeneous part I used the quadratic equation to get the roots -1 and -4.
Y = Yc + Yp
Yc = C1e^-t +C2e^-4t

I tried Yp = (At^2+Bt+C)*De^(4t) but it doesn't seem correct after solving a bunch of derivatives and plugging it back into y'' + 3y' -4y and solving for constants.

How to determine the Y-particular for this problem?

Can you show what you did and where it went wrong? The numbers aren't pretty but it certainly seems to be working fine for me. You don't need the D parameter. Set it equal to 1. Then I'll get you started. A=(-1/24).
 
Last edited by a moderator:
And one more thing, the roots of your auxiliary equation are 1 and -4 respectively.
 
sagardip said:
And one more thing, the roots of your auxiliary equation are 1 and -4 respectively.

That would be true. Nice catch. But you don't need them to find the particular solution.
 

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