Solve the differential equation dy∕dx = (y²+4)/(x²+16), y(4)=1.

Friday, August 22, 2014

Solve the differential equation dy∕dx = (y²+4)/(x²+16), y(4)=1.

Hello!

This differential equation is a separable one, it is possible to separate $\displaystyle{y}$ from $\displaystyle{x}.$ For this, simply divide both sides by $\displaystyle{\left({{y}}^{{2}}+{4}\right)}:$

$\displaystyle\frac{{{y}'}}{{{{y}}^{{2}}+{4}}}=\frac{{1}}{{{{x}}^{{2}}+{16}}}.$

$\displaystyle{y}$ is at the left side only, $\displaystyle{x}$ is at the right side only. Moreover, both sides are integrable in elementary functions:

$\displaystyle\frac{{1}}{{2}}{\arctan{{\left(\frac{{y}}{{2}}\right)}}}=\frac{{1}}{{4}}{\arctan{{\left(\frac{{x}}{{4}}\right)}}}+{C},$

or $\displaystyle{\arctan{{\left(\frac{{y}}{{2}}\right)}}}=\frac{{1}}{{2}}{\arctan{{\left(\frac{{x}}{{4}}\right)}}}+{C}.$ (1)

Now use the given boundary condition, $\displaystyle{y}{\left({4}\right)}={1},$ to find $\displaystyle{C}:$

$\displaystyle{\arctan{{\left(\frac{{1}}{{2}}\right)}}}=\frac{{1}}{{2}}{\arctan{{\left({1}\right)}}}+{C},$ or

$\displaystyle{C}={\arctan{{\left(\frac{{1}}{{2}}\right)}}}-\frac{{1}}{{2}}{\arctan{{\left({1}\right)}}}={\arctan{{\left(\frac{{1}}{{2}}\right)}}}-\frac{\pi}{{8}}.$

If we take $\displaystyle{\tan{}}$ of the both sides of (1), we obtain

$\displaystyle{y}{\left({x}\right)}={2}{\tan{{\left(\frac{{1}}{{2}}{\arctan{{\left(\frac{{x}}{{4}}\right)}}}+{\arctan{{\left(\frac{{1}}{{2}}\right)}}}-\frac{\pi}{{8}}\right)}}}.$

This is the only solution.

Ninenotes

Friday, August 22, 2014

Solve the differential equation dy∕dx = (y²+4)/(x²+16), y(4)=1.

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