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I just need for someone to check my work and suggest a better way to solve this if one exists. I can use any method but not numerical or any other iterative series approximation. The following documents my process of attempting to solve this problem. It does not appear to be separable, reducible to separable such as with an appropriate change of variable like the u-substitution $u=\tfrac{y}{x}$, or linear, or reducible to linear? But is it exact or can it be made to be exact?

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

Rearrange terms and expand then test for total differential (exactness):

$$Mdx + Ndy = 0$$

$$-ydx + (x + y^2 + 2y + 1)dy = 0$$

$$ \frac {\partial M}{\partial y} = -1, \frac {\partial N}{\partial x} = 1$$

It is not exact but can it be made to be exact with an integrating factor in x? Here $P$ is $M$ and $Q$ is $N$. $P$ and $Q$ are used to imply that ODE is not exact but is being worked on to find an integrating factor.

$$ \frac {1}{F} \frac {dF}{dx} = \frac {1}{Q} \left[ \frac {\partial P}{\partial y} - \frac {\partial Q}{\partial x} \right]$$

$$ \frac {1}{F} \frac {dF}{dx} = \frac {1}{x + y^2 + 2y + 1} \left[ -1 - 1 \right] = \frac {-2}{x + y^2 + 2y + 1}$$

This will not work because of the y-terms. Plus it would be far too complex to use. We need to find a simpler integrating factor. How about deriving one in y? Like before $P$ is $M$ and $Q$ is $N$.

$$ \frac {1}{F} \frac {dF}{dy} = \frac {1}{P} \left[ \frac {\partial Q}{\partial x} - \frac {\partial P}{\partial y} \right]$$

$$ \frac {1}{F} \frac {dF}{dy} = \frac {1}{-y} \left[ 1 - (-1) \right] = - \frac {2}{y}$$

It appears we may have made a breakthrough in finding an integrating factor in y.

$$ \int \frac {dF}{F} = -2 \int \frac {dy}{y}$$

$$ln|F| = -2 \ln|y|$$

$$F = y^{-2}$$

So far so good. Let us now multiply the rearranged ODE by $F$ and test once more for exactness. As before $P$ is $M$ and $Q$ is $N$.

$$FPdx + FQdy = 0$$

$$y^{-2}(-y)dx + y^{-2}(x + y^2 + 2y + 1)dy = 0$$

$$-\frac {1}{y}dx + \left(\frac {x}{y^2} + 1 + \frac {2}{y} + \frac {1}{y^2}\right)dy = 0$$

$$ \frac {\partial M}{\partial y} = \frac {1}{y^2}, \frac {\partial N}{\partial x} = \frac {1}{y^2}$$

The integrating factor worked. We now have exactness and can finally proceed in solving this ODE. We may begin with either term $M$ or $N$.

$$u(x,y) = \int Mdx = - \int \frac {1}{y}dx = - \frac {x}{y} + k(y)$$

$$ \frac {\partial u}{\partial y} = \frac {\partial \left[-\frac{x}{y} + k(y)\right]}{\partial y} = \frac {x}{y^2} + k(y)' = N = \frac {x}{y^2} + 1 + \frac {2}{y} + \frac {1}{y^2}$$

We are now at the home stretch. We just need to isolate and solve for $k(y)$ and insert it back into our premature solution $u(x,y)$.

$$ k(y)' = 1 + \frac {2}{y} + \frac {1}{y^2}$$

$$ k(y) = \int \left[1 + \frac {2}{y} + \frac {1}{y^2}\right]dy = y + 2\ln|y| - \frac {1}{y} + c$$

$$u(x,y) = - \frac {x}{y} + k(y) = - \frac {x}{y} + y + 2\ln|y| - \frac {1}{y} + c$$

$$y + 2\ln|y| - \frac {1}{y}(x + 1) = c$$

I believe I may have solved it but I just need someone to check my work.

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You could check by yourself your result : $$y+2\ln|y|-\frac{1}{y}(x+1)=c$$ $$x=y^2-2y\ln|y|-cy-1$$ $$dx=(2y-2\ln|y|-2-c)dy$$ Bringing it back into the ODE : $$-ydx + (x + y^2 + 2y + 1)dy = 0$$ leads to : $$-y\big((2y-2\ln|y|-2-c)dy\big) + \big((y^2-2y\ln|y|-cy-1) + y^2 + 2y + 1\big)dy = $$ after simplification $$=0$$ So, the result is correct.

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  • $\begingroup$ Thank you for checking it for me. I just wanted an independent person to do that. $\endgroup$ – Jules Manson Nov 19 '14 at 7:12
  • $\begingroup$ Don't mention it, you are welcome. $\endgroup$ – JJacquelin Nov 19 '14 at 7:15
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I think you could go faster to the solution rewriting $$\frac{dy}{dx} = \frac{y}{x + (y + 1)^2}$$ as $$\frac{dx}{dy} = \frac{x + (y + 1)^2}{y}$$ that is to say $$\frac{dx}{dy}-\frac{x}{y} =\frac{ (y + 1)^2}{y}=y+2+\frac{1}{y}$$ which gives for the homogeneous equation $x=Cy$. So, the equation write now $$y \frac{dC}{dy}=y+2+\frac{1}{y}$$ that is to say $$\frac{dC}{dy}=1+\frac{2}{y}+\frac{1}{y^2}$$ and then $$C=-\frac{1}{y}+y+2 \log (y)+K$$ So, finally $$x=K y+y^2+2 y \log (y)-1$$

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  • $\begingroup$ You are right. I knew about changing dependent variable but I had forgotten about it. Thank you for this. $\endgroup$ – Jules Manson Nov 19 '14 at 7:10
  • $\begingroup$ You are very welcome ! $\endgroup$ – Claude Leibovici Nov 19 '14 at 7:11

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