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 ^ 2 * c) . is that correct? I'll show my working out below. I worked out eta as eta =( x ^ 2 * (1) \to 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(1) I worked out dr / dx = 1 , dr/dy = zero, d/dy (s) =1/(x^2 c1) and ds/dx = (2 - 2y) /(x^ ^ 3c1). 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, |get:((2-2y)/(x^3 c1))+ (((2y) / x) + x) (1/(x ^2(1))/( 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 d/dr (s) =((2/(x^ ^ 3c1))+(1/(x c1)), I take dr to the other side and integrate: integral (ds) = integral ((2/( x ^ 3 * c(1))+ (1 / (x(1))) * dr . This gives: S = (2r/(x^3 c1)) + (r/(x c1)) + c2. I now sub in my rand 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^ ^ 2c1 c2). I make C * 1c * 2 = 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 / d * x = (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?

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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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 ^ 2 * c) . is that correct? I'll show my working out below. I worked out eta as eta =( x ^ 2 * (1) \to 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(1) I worked out dr / dx = 1 , dr/dy = zero, d/dy (s) =1/(x^2 c1) and ds/dx = (2 - 2y) /(x^ ^ 3c1). 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, |get:((2-2y)/(x^3 c1))+ (((2y) / x) + x) (1/(x ^2(1))/( 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 d/dr (s) =((2/(x^ ^ 3c1))+(1/(x c1)), I take dr to the other side and integrate: integral (ds) = integral ((2/( x ^ 3 * c(1))+ (1 / (x(1))) * dr . This gives: S = (2r/(x^3 c1)) + (r/(x c1)) + c2. I now sub in my rand 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^ ^ 2c1 c2). I make C * 1c * 2 = 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 / d * x = (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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