Katherine Walls

2021-12-27

Solve the following differential equations using the indicated method.
SEPARATION OF VARIABLES: ${x}^{2}dx+y\left(x-1\right)dy=0$

Jeffery Autrey

The given differential equation is:
${x}^{2}dx+y\left(x-1\right)dy=0$
Now separate the variables:
$-ydy=\frac{{x}^{2}}{x-1}dx$
Now integrate both side:
$-\int ydy=\int \frac{{x}^{2}}{x-1}dx$
$-\frac{{y}^{2}}{2}=\int \frac{{x}^{2}-1+1}{x-1}dx$
$-\frac{{y}^{2}}{2}=\int \frac{{x}^{2}-1}{x-1}dx+\int \frac{1}{x-1}dx$
$-\frac{{y}^{2}}{2}=\int \frac{\left(x-1\right)\left(x+1\right)}{x-1}dx+\int \frac{1}{x-1}dx$
$-\frac{{y}^{2}}{2}=\int \left(x+1\right)dx+\int \frac{1}{x-1}dx$
$-\frac{{y}^{2}}{2}=\frac{{x}^{2}}{2}+x+\mathrm{ln}|x-1|+C$
Hence the general solution of the given differential equation is:
$\frac{{y}^{2}}{2}+\frac{{x}^{2}}{2}+x+\mathrm{ln}|x-1|+C=0$
where C is an arbitrary constant.

sukljama2

${x}^{2}dx+y\left(x-1\right)dy=0⇒{x}^{2}dx=-y\left(x-1\right)dy$
We can separate the xs

karton

rewrite DE as $\left[{x}^{2}/\left(x-1\right)\right]dx+ydy=0$
$\left[\left(x+1\right)+1/\left(x-1\right)\right]dx+ydy=0$
So $\left[\left(x+1{\right)}^{2}\right]/2+\mathrm{log}\left(x-1\right)+{y}^{2}/2=C$
$⇒\left(x+1{\right)}^{2}+2\mathrm{log}\left(x-1\right)+{y}^{2}=C$
$⇒{x}^{2}+2x+2\mathrm{log}\left(x-1\right)+{y}^{2}=C-4=K$ (constant) is the solution

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