 Elleanor Mckenzie

2021-09-13

$\left(\frac{dy}{dx}\right)=-\frac{{y}^{2}+{x}^{2}}{2xy}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}y\left(1\right)=4$
Please, solve the differential equation. Write the method you used and solve for the dependent variable it it is possible. doplovif

The given equation:
${y}^{\prime }=-\frac{{y}^{2}+{x}^{2}}{2xy},y\left(1\right)=4$
The given ODE can be written as:
${y}^{\prime }=-\frac{y}{2x}-\frac{x}{2y}$
implies ${y}^{\prime }+\left(\frac{1}{2x}\right)y=\left(-\frac{x}{2}\right){y}^{-1}$ (1)
A first oredr Bernoulli ODE has the form of
${y}^{\prime }+p\left(x\right)y=q\left(x\right){y}^{n}$
In this case we have also given a first order Bernoulli ODE with
$p\left(x\right)=\frac{1}{2x},q\left(x\right)=-\frac{x}{2},n=-1$
Note that the general solution is obtained by substituting $\upsilon ={y}^{1-n}$ and solving $\frac{1}{1-n}{\upsilon }^{\prime }+p\left(x\right)\upsilon =q\left(x\right)$
Putting $\upsilon ={y}^{2}$ in (1) and using ${\upsilon }^{\prime }=2y{y}^{\prime }$ we get
${\upsilon }^{\prime }+\left(\frac{1}{x}\right)=-x$
Now by multipling the integrating factor $\mathrm{exp}\left(\int \frac{1}{x}dx\right)=\mathrm{exp}\left(\mathrm{ln}x\right)=x$ we get
$\upsilon =\frac{-{x}^{3}+{c}_{1}}{3x}$
Now substitute back $\upsilon ={y}^{2}$
${y}^{2}=\frac{-{x}^{3}+{c}_{1}}{3x}$
Apply initial conditions $y\left(1\right)=4$ gives us ${c}_{1}=49$ and so we get ${y}^{2}=\frac{-{x}^{3}+49}{3x}$
Note that $y=-\sqrt{\frac{-{x}^{3}+49}{3x}}$, not possible as it does not gives us $y\left(1\right)=4$
So, the required solution is $y=+\sqrt{\frac{-{x}^{3}+49}{3x}}$

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