Question

Solve the differential Equation:

$(1-x^2{)}^{2}dy+(y\sqrt{1-x^2}-x-\sqrt{1-x^2})dx=0$.

Answer 1

$M(x,y)=(y-1)\sqrt{1-x^2-x}$
$N(x,y)=(1-x^2{)}^{2}dy$
We can see that $\frac{d\mu }{dy}\ne \frac{dN}{dx}$
$\Rightarrow M(x,y)dx+N(x,y)dy=0$ is not a eract D. E
Let $\mu (x,y)M(x,y)dx+\mu (x,y)N(x,y)dy=0$ be an eract D. E
Where$\mu (x,y)=$integrating fraction
$\frac{1}{\mu }=\frac{\frac{d\mu }{dy}-\frac{dN}{dn}}{N\frac{d\mu }{dx}-M\frac{2\mu }{dy}}$
$\frac{d\mu }{dy}=\sqrt{1-x^2}$
$\frac{dN}{dx}=2(1-x^2)=(-2x)$
In case, the integrating factor $\mu$ depends only on $x$
$=\frac{1}{\mu }.\frac{d\mu }{dx}=\frac{\frac{d\mu }{dy}-\frac{dN}{dx}}{N}$
$=\frac{\sqrt{1-x^2}+\mu x(1-x^2)}{(1-x^2{)}^2}$
$\Rightarrow \int \frac{d\mu }{\mu }=\int \left[\right(1-x^2{)}^{-3/2}-2.\frac{-2xdx}{(1-x^2)}]dx$
$\Rightarrow ln\left(\mu \right)=\frac{-2}{\sqrt{1-x^2}}-2ln(1-x^2)$
$\Rightarrow ln\left(\mu \right(1-x^2{)}^2=\frac{-2}{\sqrt{1-x^2}}$
$\Rightarrow \mu \left(n\right)=\frac{1}{(1-x^2{)}^2}.e^{\frac{-2}{\sqrt{1-x^2}}}$
$\Rightarrow \frac{1}{(1-x^2{)}^2}=e^{\frac{-2}{\sqrt{1-x^2}}}.\left[\right(y-1)\sqrt{1-x^2}-x]dx+e^{\frac{-2}{\sqrt{1-x^2}}}dy=0$
$=f(x,y)=\int e^{\frac{-2}{\sqrt{1-x^2}}}dy+h\left(n\right)$
$=y.e^{\frac{-2}{\sqrt{1-x^2}}}+h\left(n\right)$
$\frac{df}{dx}=y.e^{\frac{-2}{\sqrt{1-x^2}}}.\frac{1}{(1-x^2{)}^2}+\frac{dh}{dx}$
$\Rightarrow \frac{dh}{dx}=\frac{e^{\frac{-2}{\sqrt{1-x^2}}}}{(1-x^2{)}^{3/2}}-\frac{x.e^{\frac{-2}{\sqrt{1-x^2}}}}{(1-x^2{)}^2}$
Intergerating with respect to $x$ ; we get
$h\left(x\right)=-x{e}^{\frac{-2}{\sqrt{1-x^2}}}+c$
$\therefore f(x,y)=(y-x).e^{\frac{-2}{\sqrt{1-x^2}}}+c$