Differential equation - question

Differential equation - question

Postby Guest » Sun May 30, 2021 7:45 am

Find a function f(x) which is derivative twice which holds the equation:

-2y'(x) = y(1/x)

have to show a mathematic way and not by guessing
Guest
 

Re: Differential equation - question

Postby Guest » Mon Jun 21, 2021 1:46 pm

Guessing (and then checking) is a valid and time honored "mathematical method"!

Now, the great majority of differential equations (in very precise sense, "almost all") have no solution in terms of "elementary functions". What reason do you have to believe this equation does?
Guest
 

Re: Differential equation - question

Postby Guest » Thu Jun 24, 2021 5:47 pm

Let z= 1/x. Then x= 1/z= z^{-1} so dx/dz= -z^{-2}. d(y(1/x))/dx= (dy(z))/dz)(dz/dx)= -z^{-2}(dy/dz)= y(z).

dy/y= -z^2 dz

ln(y)= -(1/3)z^3+ C

y(z)= e^{-(1/3)z^3+ C}= C' e^{-(1/3)z^3} where C'= e^C.

So y(x)= C'e^{-1/3x^3}
Guest
 

Re: Differential equation - question

Postby Guest » Thu Jun 24, 2021 5:48 pm

Let z= 1/x. Then x= 1/z= z^{-1} so dx/dz= -z^{-2}. d(y(1/x))/dx= (dy(z))/dz)(dz/dx)= -z^{-2}(dy/dz)= y(z).

dy/y= -z^2 dz

ln(y)= -(1/3)z^3+ C

y(z)= e^{-(1/3)z^3+ C}= C' e^{-(1/3)z^3} where C'= e^C.

So y(x)= C'e^{-1/3x^3}
Guest
 


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