4

I start out with the separable differential equation,

$$y' =\frac{dy}{dx} = \frac{y^2 + 1}{xy + y} = \frac{y^2 + 1}{y(x+1)}$$

Thus, $\frac{1 }{x+1}dx = \frac{y }{y^2 + 1}dy$.

Then integrating both sides of the equation, I get

$$\ln(x+1) = \frac{1}{2}\ln(y^2 +1) + C$$

Now, $e^{\ln(x+1)}$ = $e^{\frac{1}{2}\ln(y^2 +1) + C}$. So...

$$(x+1) = e^C(y^2 + 1)^{\frac{1}{2}}$$

I kind of wanted to know if this is indeed the correct general formula. And also, how would I find the singular solution, if there happens to be one in this case.

nmasanta
  • 9,640
Skm
  • 2,392
  • There should not be singular solution for a first order ode which is linear in $y'$. Do you know of an example? By singular solution here I mean the envelope of the family of general solution. i.e. solution that can not be obtained for any value of the constant of integration. In your example some will say that $y=\pm i$ are singular solutions. But these can be obtained from general solution. – Nasser May 04 '24 at 05:17

3 Answers3

1

You should get $x+1=C_1\sqrt{y^2+1}$, where $C_1=e^C$.

alans
  • 6,643
  • 1
  • 26
  • 51
  • That's true, I missed that. How would I solve for the singular solution, though? – Skm Feb 18 '18 at 07:14
0

You correctly obtained the general solution : $$x+1=C_1\sqrt{y^2+1} \tag 1$$ Writing it on explicit form for $y(x)$ : $$y(x)=\pm\sqrt{C_2(x+1)^2-1} \tag 2$$ with $C_2=\frac{1}{C_1^2}$

The result presented on the form $(2)$ forgets the particular case $C_1=0$ which corresponds to the singular solution $x=-1$, any $y$, that is $x(y)=-1$ , which is represented by a vertical straight line in Cartesian coordinates.

The line $x(y)=-1$ is the envelope of the curves of Eq.$(2)$.

JJacquelin
  • 68,401
  • 4
  • 40
  • 91
0

I believe that $x + 1 = \pm e^C\sqrt{y^2 + 1}$ is the correct general solution. You can rewrite it as: $x + 1 = C\sqrt{y^2 + 1}$, where $C = \pm e^c \neq 0$.

As for your concern about the singular solution, I don't think the original equation has $x = -1$ as the singular solution because $x = -1$ makes the original equation undefined. If only the original equation is initially given in the form: \begin{align}y'(x + xy) = y^2 + 1\end{align}

then the solution $x = -1$ does not make the original equation undefined and therefore is singular.

Apologize for bad Engish.

thanh.phan95
  • 31