I’m doing first order ODE questions using lie symmetries. The question is:  Dy/dx = (2y/x) + x = w(x, y)   Assume zeta = 0 and eta = eta(x).  my final answer is y = 2 + (x^2) + (x^2c). is that correct? I’ll show my working out below.   I worked out eta as eta= (x^2 c1) -> after subbing in my assumptions eta = eta(x) and zeta = 0. I subbed this into the canonical variables equations to get: r=x, s= y/(x^2 c1).   I worked out dr/dx = 1, dr/dy = zero, ds/dy = 1/(x^2 c1) and ds/dx = (2-2y)/ (x^3 c1). Omega(x,y) is given in the question. I now do ds/dr using the formula: [ds/dx + omega(x,y)ds/dy] / [ dr/dx + omega(x,y)dr/dy].   when I plug everything into the ds/dr formula, I get: ((2-2y)/ (x^3 c1)) + (((2y)/x) + x)(1/ (x^2 c1)) / (1 + (((2y)/x) + x)(0). the denominator becomes 1, and I expand the numerator: ((2-2y) / (x^3 c1)) + ((2y) / (x^3 c1)) + x / (x^2 c1). then the 2y cancels out on top to give (2 / (x^3 c1)) + (x/(x^2 c1)).  This can be simplified to (2 / (x^3 c1)) + (1/(x c1)).   Now that I have ds/dr = ((2 / (x^3 c1)) + (1/(x c1)), I take dr to the other side and integrate: integral (ds) = integral ((2 / (x^3 c1)) + (1/(x c1))) dr. This gives: S = (2r / (x^3 c1)) + (r/(x c1)) + c2. I now sub in my r and s values: r=x and s= y / (x^2 c1), to transform back to x and y ->  y / (x^2 c1) = (2x / (x^3 c1)) + (x / (x c1)) + c2. I get the y = 2 + (x^2) + (x^2 c1c2). I make c1c2=c. I get the final answer of y = 2 + (x^2) + (x^2 c). Is that correct?   Can you check my solution, y = 2 + (x^2) + (x^2 c), by substituting into the original equation, Dy/dx = (2y/x) + x. when i did it is not working. could you go through my working out and tell me where i went wrong please?

Calculus For The Life Sciences
2nd Edition
ISBN:9780321964038
Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Chapter6: Applications Of The Derivative
Section6.CR: Chapter 6 Review
Problem 20CR
Question

I’m doing first order ODE questions using lie symmetries. The question is:  Dy/dx = (2y/x) + x = w(x, y)

 

Assume zeta = 0 and eta = eta(x).  my final answer is y = 2 + (x^2) + (x^2c). is that correct? I’ll show my working out below.

 

I worked out eta as eta= (x^2 c1) -> after subbing in my assumptions eta = eta(x) and zeta = 0. I subbed this into the canonical variables equations to get: r=x, s= y/(x^2 c1).

 

I worked out dr/dx = 1, dr/dy = zero, ds/dy = 1/(x^2 c1) and ds/dx = (2-2y)/ (x^3 c1). Omega(x,y) is given in the question. I now do ds/dr using the formula: [ds/dx + omega(x,y)ds/dy] / [ dr/dx + omega(x,y)dr/dy].

 

when I plug everything into the ds/dr formula, I get: ((2-2y)/ (x^3 c1)) + (((2y)/x) + x)(1/ (x^2 c1)) / (1 + (((2y)/x) + x)(0). the denominator becomes 1, and I expand the numerator: ((2-2y) / (x^3 c1)) + ((2y) / (x^3 c1)) + x / (x^2 c1). then the 2y cancels out on top to give (2 / (x^3 c1)) + (x/(x^2 c1)).  This can be simplified to (2 / (x^3 c1)) + (1/(x c1)).

 

Now that I have ds/dr = ((2 / (x^3 c1)) + (1/(x c1)), I take dr to the other side and integrate: integral (ds) = integral ((2 / (x^3 c1)) + (1/(x c1))) dr.

This gives: S = (2r / (x^3 c1)) + (r/(x c1)) + c2.

I now sub in my r and s values: r=x and s= y / (x^2 c1), to transform back to x and y ->  y / (x^2 c1) = (2x / (x^3 c1)) + (x / (x c1)) + c2.

I get the y = 2 + (x^2) + (x^2 c1c2). I make c1c2=c.

I get the final answer of y = 2 + (x^2) + (x^2 c). Is that correct?   Can you check my solution, y = 2 + (x^2) + (x^2 c), by substituting into the original equation, Dy/dx = (2y/x) + x.

when i did it is not working. could you go through my working out and tell me where i went wrong please?

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