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Which method should I use to solve this equation ?

$$\left(x-1\right)\frac{dy}{dx}-x\left(4x+5\right)+4\left(2x+1\right)y\:-\:4y^2\:=\:0 $$

I tried using substitution method. please give me a hint.

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  • $\begingroup$ Hint: This is a Riccati type equation. $\endgroup$ – Yves Daoust Jan 11 '16 at 11:27
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The Riccati equation is easy to solve if a particular solution is known. Often, the main difficulty is to find a particular solution. But, since it is probably an exercise for student, probably a particular solution of very simple form exists. This draw to try some simple elementary functions, for example a linear function $\:y=ax+b\:$. This trial is successful : By easy identification, one find $\:a=1\:$ and $_\:b=\frac{1}{2}\:$

Then, following the usual method of change of function : $$y(x)=u(x)+x+\frac{1}{2}$$ puting it into the ODE, and after simplification : $$(x-1)u'=4u^2$$ It is easy to solve this separable ODE : $$u=-\frac{1}{4\ln(x-1)+c}$$ $$y=x+\frac{1}{2}-\frac{1}{4\ln(x-1)+c}$$

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  • $\begingroup$ You have used $$ y(x)= u(x)+x+1/2 $$, what will be the answer if I take the function as $$ y(x)= 1/u(x)+x+1/2 $$ ? $\endgroup$ – Falcon Jan 12 '16 at 13:48
  • $\begingroup$ The final result will be exactly the same. But take care of a possible confusion : do not use the same symbol for two different functions. So, better let $$y(x)=1/v(x)+x+1/2$$ . Solving the transformed ODE will give $$v(x)=4\ln(x-1)+c$$. $\endgroup$ – JJacquelin Jan 12 '16 at 14:10
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HINT:

$$(x-1)\frac{\text{d}y}{\text{d}x}-x(4x+5)+4(2x+1)y(x)-4y(x)^2=0\Longleftrightarrow$$ $$y'(x)(x-1)-x(4x+5)+4y(x)(2x+1)-4y(x)^2=0\Longleftrightarrow$$ $$y'(x)=\frac{4y(x)^2-4y(x)(1+2x)+x(5+4x)}{x-1}\Longleftrightarrow$$ $$y'(x)=\frac{4y(x)^2}{x-1}-\frac{4y(x)(1+2x)}{x-1}+\frac{x(5+4x)}{x-1}$$

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  • $\begingroup$ -1. How is this a hint, in any way? $\endgroup$ – Did Jan 12 '16 at 7:20

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