Q119 / 160 · TS EAMCET 2006 · Mathematics

The solution of $\left(1+x^2\right) \frac{d y}{d x}+2 x y-4 x^2=0$ is :

  • A
    $3 x\left(1+y^2\right)=4 y^3+c$
  • Correct answer
    B
    $3 y\left(1+x^2\right)=4 x^3+c$
  • C
    $3 x\left(1-y^2\right)=4 y^3+c$
  • D
    $3 y\left(1+y^2\right)=4 x^3+c$

Solution

$\left(1+x^2\right) \frac{d y}{d x}+2 x y-4 x^2=0$ $\Rightarrow \quad \frac{d y}{d x}+\left(\frac{2 x}{1+x^2}\right) y=\frac{4 x^2}{1+x^2}$ On comparing with $\frac{d y}{d x}+P y=Q$, we get $P=\frac{2 x}{1+x^2}, Q=\frac{4 x^2}{1+x^2}$ I.F. $=e^{\int P d x}=e^{\log \left(1+x^2\right)}$ $=1+x^2$ The solution is $y \text { (I.F.) }=\int Q \text { (I. F. ) } d x+c_1$ $\Rightarrow \quad y\left(1+x^2\right)=\int \frac{4 x^2}{\left(1+x^2\right)}\left(1+x^2\right) d x+C_1$ $\Rightarrow \quad y\left(1+x^2\right)=\frac{4 x^3}{3}+c_1$ $\Rightarrow \quad 3 y\left(1+x^2\right)=4 x^3+c$

Hence, the correct answer is option: 'B'.

Relevant Questions