Priscilla Johnston

Answered

2022-01-06

Answer & Explanation

movingsupplyw1

Expert

2022-01-07Added 30 answers

Solution:

$(4-x{y}^{2})dx+(x+{x}^{2}{y}^{2})dx=0$

$(4dx+xdy)-(x{y}^{2}dy-{x}^{2}{y}^{2})=0$

$d\left(xy\right)-{\left(xy\right)}^{2}[\frac{dx}{x}-dy]=0$

$\int \frac{d\left(xy\right)}{{\left(xy\right)}^{2}}-\int \frac{dx}{x}+\int dy=0$

$\Rightarrow -\frac{1}{x\cdot y}-\mathrm{ln}x+y=C$

put$4\left(1\right)=1$

$-\frac{1}{1\cdot 1}-\mathrm{ln}1+1=C\Rightarrow -1-0+1=C$

$c=0$

Hence,$-\frac{1}{x\cdot y}\cdot \mathrm{ln}x+y=0$

put

Hence,

SlabydouluS62

Expert

2022-01-08Added 52 answers

Consider the differential equation

$(y-x{y}^{2})dx+(x+{x}^{2}{y}^{2})dy=0$

$ydx-x{y}^{2}dx+xdy+{x}^{2}{y}^{2}dy=0$

$(ydx+xdy)+({x}^{2}{y}^{2}dy-x{y}^{2}dx)=0$

$d\left(xy\right)+xy(xydy-ydx)=0$

$d\left(xy\right)+{x}^{2}{y}^{2}dy-x{y}^{2}dx=0$

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$\int \frac{dz}{{z}^{2}}=-\frac{1}{z}$

$\int {x}^{n}dx={x}^{n}+\frac{1}{n+1}+c\text{}\text{for}\text{}n\ne -1$

$\int \frac{dx}{x}=\mathrm{ln}x+c$

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.

Dividing by

Integrating the equation above we get

Where C is constant of integration

Now say

Using these identities if

Multiply throughout by xy in the above equation we get

Therefore the solution of the equation is

Where C is any arbitrary constant.

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