Question

Solve the given differential equation :

(2xy + x²) dy/dx = 3y² + 2xy

please help. ​

Answer 1

Solve Given equation $2xy \frac{dy}{dx} + x^{2}\frac{dy}{dx} -3y^{2} -2xy= 0$

Step-by-step explanation:

1. do substitution,        $y(x)= x.u(x)$                  -----(a)                                                                                              differentiating it we get,     $\frac{dy}{dx} = x\frac{du}{dx} + u(x)$              

    2. substituting this in the given differential equation we get,    

                         $3x^{2} u^{2} - 2x^{2} u\frac{d(xu)}{dx} + 2x^{2} u-x^{2} \frac{d(xu)}{dx} =0\\x^{2} u^{2}+x^{2} u-2x^{3} u\frac{du}{dx} - x^{3} \frac{du}{dx}=0$                                                                                                                                

                          $\frac{du}{dx} = \frac{u(u+1)}{x(2u+1)}$                         -----(i)

     3. above equation (i) is of the form , $f_1(x).g_1(u).\frac{du}{dx} = f_2(x).g_2(u)$                                  

         here,   $f_1(x)= g_1(u)= 1 \\f_2(x)= -\frac1{x}\ and\ g_2(u)= -\frac{u(u+1)}{2u+1}$        

     4. separating both variables x and u in equation (i)  and integrating     them we get,

                        $\int\ \frac{2u+1}{u(u+1)} \, du= \int\frac1 {x} \, dx$

      5. solving above integration we get,    $log(u^2 + u) = log(x) + c$

      6. solving for u(x) we get,   $u(x)= (\frac{-\sqrt{c'+1} -1 }{2} ),(\frac{\sqrt{c'+1} -1 }{2})$

      7. putting these values of u(x) in (a) we get our solution ,

                        $y(x)= (x\frac{-\sqrt{c'+1} -1 }{2} ),(x\frac{\sqrt{c'+1} -1 }{2})$            (here c' is a constant)