First-Order Differential Equation with Homogeneous Coefficients

[Mr. Silva]

I am solving the equation \(\displaystyle \, \left(2x^2y\, +\, y^3\right) dx\, +\, \left(xy^2\, -\, 2x^3\right) dy\, =\, 0\,\) by using the substitutions \(\displaystyle \,x\,=\,uy\,\) and \(\displaystyle \, dx\, =\, ydu\, +\, udy.\,\) Doing this gives me the implicit solution:

. . .\(\displaystyle \ln\Bigg|\, xy\, \Bigg|\, +\, \left(\dfrac{x}{y}\right)^2\, =\, c\)

My answer key gives me the same solution without the absolute value bars, but I can't figure out how I can drop them.

Thanks for reading.

 

[alain]

I am solving the equation \(\displaystyle \, \left(2x^2y\, +\, y^3\right) dx\, +\, \left(xy^2\, -\, 2x^3\right) dy\, =\, 0\,\) by using the substitutions \(\displaystyle \,x\,=\,uy\,\) and \(\displaystyle \, dx\, =\, ydu\, +\, udy.\,\) Doing this gives me the implicit solution:

. . .\(\displaystyle \ln\Bigg|\, xy\, \Bigg|\, +\, \left(\dfrac{x}{y}\right)^2\, =\, c\)

My answer key gives me the same solution without the absolute value bars, but I can't figure out how I can drop them.

Thanks for reading.

You can drop the absolute value bars because there is no solution to this implicit equation where x and y are not of the same sign. The equation can be rewritten as xy=e^(-(x/y)^2+c). The right hand side of this equation is always >0, so x and y are always of the same sign, that is xy always > 0, therefore no need for absolute value bars.