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HallsofIvy

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## Main Question or Discussion Point

gerechte23 added this to a thread on a different problem. I am starting a new thread for it

Hold on, I was assuming that the "x" in your equation was the independent variable. Did you mean it simply as "times"? (unlikely- by the time you are doing differential equations you should know better!

IF that was what you meant, then since the independent variable does not appear explicitely, we can use "quadrature". Let v= Y'. Then Y"= V'= dV/dx= (dV/dY)(dY/dx)= V(dV/dY) so the equation becomes the separable first order equation Y^2V dV/dY= C which can be written V dV= CdY/Y^2 and integrates to [itex](1/2)V^2= -C/Y+ D[/itex] where D is the constant of integration. Now, since V= Y', we can write that as the first order equation (1/2)(Y')^2= -C/Y+ D= (DY- C)/Y or Y'= 2(DY- C)/Y, again a separable equation.

[tex]\frac{YdY}{DY- C}= 2dx[/tex]

To integrate the left side, let u= DY- C so that du= Ddy and we have

[tex]\frac{1}{D}\frac{du}{u}= 2dx[/tex]

That is easy to integrate.

Of course, that only works if you meant Y

Why do you think you ought to be able to resolve it? That looks to me like a very nasty non-linear equation.--------------------------------------------------------------------------------

Hello, I am calculating a very important thing. But to find it, I should be able to resolve this equation. Y^2xY’’=C (C: Real number). Please help me solve it. Thanks in advance!!

(YY)xY''=C (C:Real nnumber)

Hold on, I was assuming that the "x" in your equation was the independent variable. Did you mean it simply as "times"? (unlikely- by the time you are doing differential equations you should know better!

IF that was what you meant, then since the independent variable does not appear explicitely, we can use "quadrature". Let v= Y'. Then Y"= V'= dV/dx= (dV/dY)(dY/dx)= V(dV/dY) so the equation becomes the separable first order equation Y^2V dV/dY= C which can be written V dV= CdY/Y^2 and integrates to [itex](1/2)V^2= -C/Y+ D[/itex] where D is the constant of integration. Now, since V= Y', we can write that as the first order equation (1/2)(Y')^2= -C/Y+ D= (DY- C)/Y or Y'= 2(DY- C)/Y, again a separable equation.

[tex]\frac{YdY}{DY- C}= 2dx[/tex]

To integrate the left side, let u= DY- C so that du= Ddy and we have

[tex]\frac{1}{D}\frac{du}{u}= 2dx[/tex]

That is easy to integrate.

Of course, that only works if you meant Y

^{2}Y"= C. If your really meant xY^{2}Y"= C with x the independent variable, I'm afraid it cannot be integrated in terms of elementary functions.