How Do You Solve Differential Equations with Variable Power Terms?

  • MHB
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In summary, in problem B, the equations \[y_1 = c_1x^2\] and \[y_2 = c_2x^{-2}\] correspond to \[y_1y_2 = c_1c_2 = c = constant\], and this allows us to find the derivatives \[y_1', y_1'', y_2', y_2''\] by solving the equation \[2p_1p_2 + p_2' = 0\]. However, in problem C, we do not have a clear solution, so we assign \[y = c_1x^r + c_2x^s\] and
  • #1
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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  • #2
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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