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


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

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Martin R, Sahiba Arora, TheGeekGreek, C. Falcon, Foobaz John
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 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

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 $\;t=y^{1/2}\implies dt=\frac{dy}{2\sqrt y}\implies dy=2tdt\implies\;$



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

  • 1
    $\begingroup$ Great answer. Thank you! $\endgroup$ – AAli Jan 6 '18 at 21:21

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 $x$.


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 $$y'(x)=\frac{1}{1+\frac{1}{\sqrt{y(x)}}-y(x)^{3/2}}-\frac{\sqrt{y(x)}}{1+\frac{1}{\sqrt{y(x)}}-y(x)^{3/2}}-\frac{y(x)}{1+\frac{1}{\sqrt{y(x)}}-y(x)^{3/2}}$$

  • $\begingroup$ 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
  • 1
    $\begingroup$ @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

Not the answer you're looking for? Browse other questions tagged or ask your own question.