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Hi I'm having a slight issue trying to solve this differential equation. I would greatly appreciate any help or tips to solve this problem.

enter image description here

click here for the equation please

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closed as off-topic by Martin R, Sahiba Arora, TheGeekGreek, C. Falcon, Foobaz John Jan 6 '18 at 23:26

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  • 2
    $\begingroup$ AAli could you please type out your question rather than insert an image, thanks. Also, could you let us know what you've tried thus far? $\endgroup$ – Colm Bhandal Jan 6 '18 at 20:59
  • $\begingroup$ Please use MathJax (i.e. LaTeX commands) to format mathematical notations, see math.stackexchange.com/help/notation $\endgroup$ – Taroccoesbrocco Jan 6 '18 at 21:12
  • $\begingroup$ Use $y=u^2$ and $dy=2udu$ to simplify the equation $\endgroup$ – Mostafa Ayaz Jan 6 '18 at 21:22
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Well, you can by sure separate it...:

$$\frac{1+\sqrt y-\frac1y}{\sqrt y-y-y\sqrt y}dy=dx\implies-\int\frac{y^{3/2}+y-1}{y^{5/2}+y^2-y^{3/2}}dy=\int dx=x+C$$

substitute in the left integral $\;t=y^{1/2}\implies dt=\frac{dy}{2\sqrt y}\implies dy=2tdt\implies\;$

$$-\int\frac{y^{3/2}+y-1}{y^{5/2}+y^2-y^{3/2}}dy=-\int\frac{t^3+t^2-1}{t^5+t^4-t^3}2t\,dt=-2\int\frac{t^4+t^3-t}{t^3(t^2+t-1)}=$$

$$=-2\int\frac{t^4+t^3-t}{t^3\left(t+\frac{1+\sqrt5}2\right)\left(t+\frac{1-\sqrt5}2\right)}dt$$

Now, the integral is ugly as one can expect it to be...but it is solvable by simple fractions.

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    $\begingroup$ Great answer. Thank you! $\endgroup$ – AAli Jan 6 '18 at 21:21
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General method. $$ \frac{dy}{dx} = G(y) \\ \frac{dx}{dy} = \frac{1}{G(y)} \\ x = \int \frac{1}{G(y)}\;dy + C $$ and solve the result for $y$ in terms of $x$.

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this equation is separable $$\frac{y'(x)}{\frac{1-\sqrt{y(x)}-y(x)}{1+\frac{1}{\sqrt{y(x)}}-y(x)^{3/2}}}=1$$ and then $$y'(x)=\frac{1}{1+\frac{1}{\sqrt{y(x)}}-y(x)^{3/2}}-\frac{\sqrt{y(x)}}{1+\frac{1}{\sqrt{y(x)}}-y(x)^{3/2}}-\frac{y(x)}{1+\frac{1}{\sqrt{y(x)}}-y(x)^{3/2}}$$

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  • $\begingroup$ why the $-1$? i reelly don't understand $\endgroup$ – Dr. Sonnhard Graubner Jan 6 '18 at 21:15
  • $\begingroup$ The second line is identical to the original equation, I cannot see how that helps towards a solution via separation of variables. Am I overlooking something? $\endgroup$ – Martin R Jan 6 '18 at 21:22
  • $\begingroup$ this line is the hint how we can integrate the term, you gave me the $-1$? $\endgroup$ – Dr. Sonnhard Graubner Jan 6 '18 at 21:24
  • $\begingroup$ yes you are overlooking the main step $\endgroup$ – Dr. Sonnhard Graubner Jan 6 '18 at 21:24
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    $\begingroup$ @Dr.SonnhardGraubner I've no idea who downvoted (some people around here has a very quick finger on the mouse...), but for remarking that the equation is separable your answer doesn't seem to make things much simpler for the asker... $\endgroup$ – DonAntonio Jan 6 '18 at 21:24

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