arenceabigns

2021-03-07

Solve differential equation $\left(1+{y}^{2}+x{y}^{2}\right)dx+\left({x}^{2}y+y+2xy\right)dy=0$

estenutC

First Order Bernoulli differential equation is of the form ${y}^{\prime }+p\left(x\right)y=q\left(x\right){y}^{n}$

Put $\frac{dy}{dx}={y}^{\prime }$
(1)
Now converting the above equation (1) in the form
${y}^{\prime }+p\left(x\right)y=q\left(x\right){y}^{n}$
Dividing the whole equation (1) by $\left({x}^{2}y+y+2xy\right)$
$\left(\frac{1+{y}^{2}+x{y}^{2}}{{x}^{2}y+y+2xy}+\frac{{x}^{2}y+y+2xy}{{x}^{2}y+y+2xy}\right){y}^{\prime }=0$
$\left(\frac{1+{y}^{2}\left(x+1\right)}{y\left({x}^{2}+1+2x\right)}\right)+{y}^{\prime }=0$
$\frac{1+{y}^{2}\left(x+1\right)}{y\left(x+1{\right)}^{2}}+{y}^{\prime }=0$
$\frac{1}{y\left(x+1{\right)}^{2}}+\frac{{y}^{2}\left(x+1\right)}{y\left(x+1{\right)}^{2}}+{y}^{\prime }=0$
$\frac{1}{x+1}y+{y}^{\prime }=-{\frac{1}{x+1}}^{2}{y}^{-1}$
${y}^{\prime }+\frac{1}{x+1}y=-{\frac{1}{x+1}}^{2}{y}^{-1}$
$p\left(x\right)=\frac{1}{x+1}$
$q\left(x\right)=-{\frac{1}{x+1}}^{2}$
$n=-1$ Now, substitute $v={y}^{1-n}$ in the Bernoulli equation, we get
$v={y}^{1-n}$
$y={v}^{\frac{1}{1}-n}$
$\frac{1}{1-n}{v}^{\prime }+p\left(x\right)v=q\left(x\right)$
Solving by putting all the values we get
$\frac{{v}^{\prime }}{2}+\frac{v}{x+1}=-{\frac{1}{x+1}}^{2}$
Put $v={y}^{2}$
and then solve for v
$v=-\frac{2x}{{x}^{2}+1+2x}+\frac{{c}_{1}}{{x}^{2}+1+2x}$
Substituting back the value of y
$y=\sqrt{-\frac{2x}{{x}^{2}+1+2x}+\frac{{c}_{1}}{{x}^{2}+1+2x}}$
$y=\sqrt{-2x+\frac{{c}_{1}}{{x}^{2}+2x+1}}$

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