vegetatzz8s

2022-11-25

How can I solve this differential equation? : $xydx-({x}^{2}+1)dy=0$

SquingonottokQqk

Beginner2022-11-26Added 8 answers

we have in differential form:

$xydx-({x}^{2}+1)dy=0$

If we put in standard form, and collect terms:

$\frac{1}{y}y\frac{dy}{dx}=\frac{x}{{x}^{2}+1}$

Which is a First Order Separable Ordinary Differential Equation, so we can separate the variables to get:

$\int \frac{1}{y}dy=\int \frac{x}{{x}^{2}+1}dx$

We can manipulate the RHS integral as follows:

$\int \frac{1}{y}dy=\frac{1}{2}\int \frac{2x}{{x}^{2}+1}dx$

And now both integrals are standard results, so integrating give us:

$\mathrm{ln}\left|y\right|=\frac{1}{2}\mathrm{ln}|{x}^{2}+1|+C$

Noting that we require areal solution, and writing $C=\mathrm{ln}A$, we get:

$\mathrm{ln}y=\mathrm{ln}A\sqrt{{x}^{2}+1}$

Giving us the General Solution:

$y=A\sqrt{{x}^{2}+1}$

$xydx-({x}^{2}+1)dy=0$

If we put in standard form, and collect terms:

$\frac{1}{y}y\frac{dy}{dx}=\frac{x}{{x}^{2}+1}$

Which is a First Order Separable Ordinary Differential Equation, so we can separate the variables to get:

$\int \frac{1}{y}dy=\int \frac{x}{{x}^{2}+1}dx$

We can manipulate the RHS integral as follows:

$\int \frac{1}{y}dy=\frac{1}{2}\int \frac{2x}{{x}^{2}+1}dx$

And now both integrals are standard results, so integrating give us:

$\mathrm{ln}\left|y\right|=\frac{1}{2}\mathrm{ln}|{x}^{2}+1|+C$

Noting that we require areal solution, and writing $C=\mathrm{ln}A$, we get:

$\mathrm{ln}y=\mathrm{ln}A\sqrt{{x}^{2}+1}$

Giving us the General Solution:

$y=A\sqrt{{x}^{2}+1}$

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