Priscilla Johnston

2022-01-06

movingsupplyw1

Expert

Solution:
$\left(4-x{y}^{2}\right)dx+\left(x+{x}^{2}{y}^{2}\right)dx=0$
$\left(4dx+xdy\right)-\left(x{y}^{2}dy-{x}^{2}{y}^{2}\right)=0$
$d\left(xy\right)-{\left(xy\right)}^{2}\left[\frac{dx}{x}-dy\right]=0$
$\int \frac{d\left(xy\right)}{{\left(xy\right)}^{2}}-\int \frac{dx}{x}+\int dy=0$
$⇒-\frac{1}{x\cdot y}-\mathrm{ln}x+y=C$
put $4\left(1\right)=1$
$-\frac{1}{1\cdot 1}-\mathrm{ln}1+1=C⇒-1-0+1=C$
$c=0$
Hence, $-\frac{1}{x\cdot y}\cdot \mathrm{ln}x+y=0$

SlabydouluS62

Expert

Consider the differential equation
$\left(y-x{y}^{2}\right)dx+\left(x+{x}^{2}{y}^{2}\right)dy=0$
$ydx-x{y}^{2}dx+xdy+{x}^{2}{y}^{2}dy=0$
$\left(ydx+xdy\right)+\left({x}^{2}{y}^{2}dy-x{y}^{2}dx\right)=0$
$d\left(xy\right)+xy\left(xydy-ydx\right)=0$
$d\left(xy\right)+{x}^{2}{y}^{2}dy-x{y}^{2}dx=0$
Dividing by ${x}^{2}{y}^{2}$
$d\frac{xy}{{x}^{2}{y}^{2}}+dy-\frac{dx}{x}=0$
$d\frac{xy}{{\left(xy\right)}^{2}}+dy-\frac{dx}{x}=0$
Integrating the equation above we get
$\int \left[d\frac{xy}{{\left(xy\right)}^{2}}\right]+\int dy-\int \frac{dx}{x}=C$…..(1)
Where C is constant of integration
Now say $\int \frac{dz}{{z}^{2}}=-\frac{1}{z}$

$\int \frac{dx}{x}=\mathrm{ln}x+c$
Using these identities if $z=xy$ then the above eqn.(1) becomes
$-\frac{1}{xy}+y-\mathrm{ln}x=C$
Multiply throughout by xy in the above equation we get
$-1+x{y}^{2}-xy\mathrm{ln}x=Cxy$
Therefore the solution of the equation is
$x{y}^{2}-xy\mathrm{ln}x-1=Cxy$.
Where C is any arbitrary constant.

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