Non linear problem

[Takitoc]

Hi,

Could someone please help me solve the following:

y = xy' + 1/y' with boundary condition y(0) = 0

Ideally the method of solution involves the following step:

Let p = y' and y = f(x,p)

Thank you,
Takitoc

 

[stapel]

y = xy' + 1/y' with boundary condition y(0) = 0

Ideally the method of solution involves the following step:

Let p = y' and y = f(x,p)

Why "ideally"? Did the instructions tell you to use this step?

What have you tried so far? Where are you stuck?

Please be complete. Thank you!


Note: Wolfram Alpha suggests that "the solution", without the boundary condition, will be along the following lines:

. . .\(\displaystyle y(x)\, =\, -\frac{
\left(
\sqrt{
x^2 \sinh(c_1)\,+\,x^2 \cosh(c_1)\, -\, 8x \sinh(c_1)\, +\, 2x \sinh(2 c_1)\, -\, 8x \cosh(c_1)\, +\, 2x \cosh(2 c_1)\, +\, 16 \sinh(c_1)\, -\, 8 \sinh(2 c_1)\, +\, \sinh(3 c_1)\, +\, 16 \cosh(c_1)\, -\, 8 \cosh(2 c_1)\, +\, \cosh(3 c_1)\,
}
\right)
}{
\left(
\sinh(c_1)\,+\,\cosh(c_1)\,-\,4
\right)
} \)

. . . . .\(\displaystyle +\, \frac{2 x}{\sinh(c_1)\,+\,\cosh(c_1)-4} \, -\, \frac{2 \cosh(c_1)}{\sinh(c_1)\,+\,\cosh(c_1)-4} \,-\, \frac{2 \sinh(c_1)}{\sinh(c_1)\,+\,\cosh(c_1)\,-\,4} \,+\, \frac{8}{\sinh(c_1)\,+\,\cosh(c_1)\,-\,4}\)

With the boundary condition, the solution cannot be found within the allotted time. Are you sure you entered the equation correctly?

 

[HallsofIvy]

If we write p for y', the equation becomes \(\displaystyle y= xp+ 1/p\). We can solve for p by first multiplying throught by p- \(\displaystyle yp= xp^2+ 1\) or \(\displaystyle xp^2- yp+ 1= 0\). Now use the quadratic formula: \(\displaystyle y'= p= \frac{y\pm\sqrt{y^2- 4x}}{2}\). But I don't see any elementary way of solving that equation.