Is This Function a Solution to the Differential Equation?

[Ammar Kurd]

Hello everyone, this is my first post in "Physics Forums"...

I need help with this problem:

Verify that y = (x^2/2) + ((x/2)*√(x^2 + 1)) + ln(√(x+√(x^2 +1))) is a solution of the

equation 2y = x*y' + ln(y')...(1)

I differentiated (y) with respect to (x), and substituted (y' and y) in equation (1), but that

led me to nowhere.

*The problem might be easy, but I study by myself and have no one to consult, I appreciate

any tips or hints, thanks in advance.

 

[tiny-tim]

Welcome to PF!

Hello Ammar Kurd! Welcome to PF!

(try using the X² button just above the Reply box )

Verify that y = (x²/2) + ((x/2)*√(x² + 1)) + ln(√(x+√(x² +1))) is a solution of the

equation 2y = x*y' + ln(y')...(1)

Show us what you got for y'

 

[Ammar Kurd]

Thank you for the replay, I got

y' = x + 0.5√(x²+1) + (x² / (2√(x²+1))) + (x+√(x²+1) / (x√(x²+1)+x²+1))

I also tried to simplify the (y) in this way:

y = 0.5x² + 0.5x√(x²+1) + ln(√(x+√(x²+1)))

= .5x(x+√(x²+1)) + 0.5ln(x+√(x²+1))

then putting G(x) = x+√(x²+1)

y becomes:

y = 0.5x*G(x) + 0.5ln(G(x)), and

y' = 0.5*G(x) +0.5xG'(x) + 0.5*(G'(x)/G(x))

But when I substitute in the differential equation it only get complicated...

*Thank you for the x² tip .

 

[tiny-tim]

Hello Ammar Kurd!

y = (x^2/2) + ((x/2)*√(x^2 + 1)) + ln(√(x+√(x^2 +1)))

Thank you for the reply, I got

y' = x + 0.5√(x²+1) + (x² / (2√(x²+1))) + (x+√(x²+1) / (x√(x²+1)+x²+1)) …

Yes, that's correct, except I think there should be a factor 2 in the last term.

I don't know how you got that last term, but it simplifies, to 1/√(x²+1)

 

[Ammar Kurd]

I don't know how you got that last term, but it simplifies, to 1/√(x²+1)

That was my problem I didn't notice that the last term can be further simplified .

Thank you, Problem solved