How Do You Solve This Challenging ODE Involving a Constant?

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SUMMARY

The discussion focuses on solving the ordinary differential equation (ODE) given by the expression \(\frac{dy}{dx} = \frac{x}{(k-y) + \sqrt{x^2 + (k-y)^2}}\), where "k" is a constant. A suggested approach involves changing variables by letting \(u = k - y\) and recognizing the equation's homogeneity of degree 1. The solution process includes substituting \(u = xv\), simplifying the equation, separating variables, integrating, and reverting back to the original variables.

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bomba923
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I'm a HS senior, and have never covered differential equations outside a CalculusII curriculum. :frown: This is not homework, rather only a part (just one part!) of an interesting problem I posed myself (for myself) one day:

How would I solve for [tex]y(x)[/tex] from :shy::

[tex]\frac{{dy}}{{dy}} = \frac{x}{{f - y + \sqrt {x^2 + \left( {f - y} \right)^2 } }}[/tex]

where "f" is a constant :smile:
 
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bomba923 said:
I'm a HS senior, and have never covered differential equations outside a CalculusII curriculum. :frown: This is not homework, rather only a part (just one part!) of an interesting problem I posed myself (for myself) one day:

How would I solve for [tex]y(x)[/tex] from :shy::

[tex]\frac{{dy}}{{dy}} = \frac{x}{{f - y + \sqrt {x^2 + \left( {f - y} \right)^2 } }}[/tex]

where "f" is a constant :smile:

Hello Bomba. May I suggest you use standard nomenclature: Reserve the 'f' for a function and just use 'k' for a constant:

[tex]\frac{dy}{dx}=\frac{x}{(k-y)+\sqrt{x^2+(k-y)^2}}[/tex]

Note that the (k-y) expression occurrs twice. How about using a change of variables for starters:

[tex]u=k-y[/tex]

Do that, then note that even though you have a square root, the equation is still homogeneous of degree 1. Such equations are normally solved by letting:

[tex]u=xv[/tex]

Make that substitution, do the algebra, simplify, separate variables, integrate, switch back to u, then back to y.:smile:
 

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