# How can I solve first order differential quation

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.

• Hint: This is a Riccati type equation. – Yves Daoust Jan 11 '16 at 11:27

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 : $$u=-\frac{1}{4\ln(x-1)+c}$$ $$y=x+\frac{1}{2}-\frac{1}{4\ln(x-1)+c}$$
• You have used $$y(x)= u(x)+x+1/2$$, what will be the answer if I take the function as $$y(x)= 1/u(x)+x+1/2$$ ? – Falcon Jan 12 '16 at 13:48
• 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 $$v(x)=4\ln(x-1)+c$$. – JJacquelin Jan 12 '16 at 14:10
$$(x-1)\frac{\text{d}y}{\text{d}x}-x(4x+5)+4(2x+1)y(x)-4y(x)^2=0\Longleftrightarrow$$ $$y'(x)(x-1)-x(4x+5)+4y(x)(2x+1)-4y(x)^2=0\Longleftrightarrow$$ $$y'(x)=\frac{4y(x)^2-4y(x)(1+2x)+x(5+4x)}{x-1}\Longleftrightarrow$$ $$y'(x)=\frac{4y(x)^2}{x-1}-\frac{4y(x)(1+2x)}{x-1}+\frac{x(5+4x)}{x-1}$$