MHB How Do You Solve Differential Equations with Variable Power Terms?

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The discussion focuses on solving differential equations with variable power terms. In problem B, the relationship \( y_1y_2 = c \) is established with \( y_1 = c_1x^2 \) and \( y_2 = c_2x^{-2} \), leading to the derivation of first and second derivatives for both functions. For problem C, the approach involves assigning \( y = c_1x^r + c_2x^s \) and deriving equations for \( y_1 \) and \( y_2 \) based on their respective second derivatives. The resulting equations are set to zero, allowing for the determination of the parameters \( r \) and \( s \). The discussion highlights the process of applying variable power terms in differential equations effectively.
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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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