Jerold

2021-02-09

Solve differential equation $\left({y}^{2}+1\right)dx=y{\mathrm{sec}}^{2}\left(x\right)dy$$y\left(0\right)=0$

### Answer & Explanation

Tasneem Almond

$\left({y}^{2}+1\right)=\left(y{\mathrm{sec}}^{2}x\right)dy/dx$
$\left({y}^{2}+1\right)=\left(y{\mathrm{sec}}^{2}x\right){y}^{\prime }$
$y/\left(-{y}^{2}-1\right){y}^{\prime }=\left(-1\right)/{\mathrm{sec}}^{2}x$
From the above, it is observed that the first order differential is in the form of $N\left(x\right){y}^{\prime }=M\left(x\right)$, where $N\left(x\right)=y/\left(-{y}^{2}-1\right)$ and $M\left(x\right)=\left(-1\right)/{\mathrm{sec}}^{2}x$
Integrate on both sides of $y/\left(-{y}^{2}-1\right)dy=\left(-1\right)/{\mathrm{sec}}^{2}xdx$
$⇒\int y/\left(-{y}^{2}-1\right)dy=\int \left(-1\right)/{\mathrm{sec}}^{2}xdx$
$⇒-\int y/\left(-{y}^{2}-1\right)dy=-\int 1/{\mathrm{sec}}^{2}xdx$
$⇒-1/2\mathrm{ln}\left(-{y}^{2}-1\right)=/1/2\left(x+1/2\mathrm{sin}\left(2x\right)\right)+c$
$-1/2\mathrm{ln}\left(0-1\right)=-1/2\left(0+1/2\mathrm{sin}\left(0\right)\right)+c$ $\left[{:}^{\prime }y\left(0\right)=0\right]$
$c=0$ Substitute $c=0\in -1/2\mathrm{ln}\left(-{y}^{2}-1\right)±1/2\left(x+1/2\mathrm{sin}\left(2x\right)\right)+c$ and solve for y.
$-1/2\mathrm{ln}\left(-{y}^{2}-1\right)=-1/2\left(x+1/2\mathrm{sin}\left(2x\right)\right)$
$\mathrm{ln}\left(-{y}^{2}-1\right)=x+1/2\mathrm{sin}\left(2x\right)$
$-{y}^{2}-1={e}^{x+1/2\mathrm{sin}\left(2x\right)}$
$-{y}^{2}={e}^{x+1/2\mathrm{sin}\left(2x\right)}+1$
${y}^{2}={e}^{x+1/2\mathrm{sin}\left(2x\right)}-1$
$y=±\sqrt{{e}^{x+1/2\mathrm{sin}\left(2x\right)}-1}$

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