(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

Elleanor Mckenzie 2021-09-13 Answered
\(\displaystyle{\left(\frac{{\left.{d}{y}\right.}}{{\left.{d}{x}\right.}}\right)}=-\frac{{{y}^{{2}}+{x}^{{2}}}}{{{2}{x}{y}}}{\quad\text{and}\quad}{y}{\left({1}\right)}={4}\)
Please, solve the differential equation. Write the method you used and solve for the dependent variable it it is possible.

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Expert Answer

doplovif
Answered 2021-09-14 Author has 18491 answers

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 \(\displaystyle{y}=+\sqrt{{{\frac{{-{x}^{{3}}+{49}}}{{{3}{x}}}}}}\)

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