How Do You Solve Differential Equations with Variable Power Terms?

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SUMMARY

This discussion focuses on solving differential equations with variable power terms, specifically addressing problems involving the product of two functions, \( y_1y_2 = c \). The user successfully defines \( y_1 = c_1x^2 \) and \( y_2 = c_2x^{-2} \) to derive the necessary derivatives and solve the equation \( 2p_1p_2 + p_2' = 0 \). In a more complex problem, the user attempts to solve \( (x+1)x^2y_1'' + xy_1' +(x+1)^3y_1 = 0 \) and \( (x+1)x^2y_2'' + xy_2' +(x+1)^3y_2 = 0 \) by assigning \( y = c_1x^r + c_2x^s \) and deriving two equations for \( r \) and \( s \).

PREREQUISITES
  • Understanding of differential equations
  • Familiarity with variable power terms
  • Knowledge of derivatives and their applications
  • Ability to manipulate algebraic expressions
NEXT STEPS
  • Study methods for solving non-homogeneous differential equations
  • Learn about the Wronskian and its role in linear independence
  • Explore the method of undetermined coefficients for variable power terms
  • Investigate the use of series solutions for differential equations
USEFUL FOR

Mathematicians, engineering students, and anyone involved in solving complex differential equations, particularly those with variable power terms.

Another1
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in problem b from \[ y_1y_2 = c \] so I was able to specify that \[ y_1 = c_1x^2 \] abd \[ y_2 = c_2x^{-2} \]
Correspond to \[ y_1y_2 = c_1c_2 = c = constant \] then I can find \[ y_1', y_1'', y_2',y_2'' \]
So. I can solve \[2p_1p_2 +p_2' = 0\]

But in problem C, I have no idea, so I assign \[ y = c_1x^r + c_2x^s \] but i can solve it
 
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I select \[ y_1= c_1x^r\] and \[ y_2= c_2x^s\]
so \[ (x+1)x^2y_1'' + xy_1' +(x+1)^3y_1 = 0\] and
\[ (x+1)x^2y_2'' + xy_2' +(x+1)^3y_2 = 0\] i find first and second deriative of y1 and y2
I get two equations
\[r(r-1)(x+1) +rx - (x+1)^3 = 0 \] \[s(s-1)(x+1) +sx - (x+1)^3 = 0 \]
 

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