# (dy/dx) = -(y^2 + x^2) / (2xy) and y(1) = 4 Write the method you used and solve for the dependent variable it it is possible

$$\displaystyle{\left(\frac{{\left.{d}{y}\right.}}{{\left.{d}{x}\right.}}\right)}=-\frac{{{y}^{{2}}+{x}^{{2}}}}{{{2}{x}{y}}}{\quad\text{and}\quad}{y}{\left({1}\right)}={4}$$
Please, solve the differential equation. Write the method you used and solve for the dependent variable it it is possible.

• Questions are typically answered in as fast as 30 minutes

### Plainmath recommends

• Get a detailed answer even on the hardest topics.
• Ask an expert for a step-by-step guidance to learn to do it yourself.

doplovif

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