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

  • Thread starter Thread starter Another1
  • Start date Start date
Another1
Messages
39
Reaction score
0
Des2.png


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
 
Physics news on Phys.org
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
\[r(r-1)(x+1) +rx - (x+1)^3 = 0 \] \[s(s-1)(x+1) +sx - (x+1)^3 = 0 \]
 
I have the equation ##F^x=m\frac {d}{dt}(\gamma v^x)##, where ##\gamma## is the Lorentz factor, and ##x## is a superscript, not an exponent. In my textbook the solution is given as ##\frac {F^x}{m}t=\frac {v^x}{\sqrt {1-v^{x^2}/c^2}}##. What bothers me is, when I separate the variables I get ##\frac {F^x}{m}dt=d(\gamma v^x)##. Can I simply consider ##d(\gamma v^x)## the variable of integration without any further considerations? Can I simply make the substitution ##\gamma v^x = u## and then...

Similar threads

  • · Replies 2 ·
Replies
2
Views
3K
Replies
2
Views
1K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 3 ·
Replies
3
Views
4K
Replies
2
Views
3K
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 15 ·
Replies
15
Views
1K
  • · Replies 3 ·
Replies
3
Views
3K