(y-xy^{2})dx + (x + x^{2}y^{2})dy = 0; \text{when}\ y(1) =

Priscilla Johnston 2022-01-06 Answered
\(\displaystyle{\left({y}-{x}{y}^{{{2}}}\right)}{\left.{d}{x}\right.}+{\left({x}+{x}^{{{2}}}{y}^{{{2}}}\right)}{\left.{d}{y}\right.}={0};\text{when}\ {y}{\left({1}\right)}={1}\)

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Expert Answer

movingsupplyw1
Answered 2022-01-07 Author has 5411 answers
Solution:
\(\displaystyle{\left({4}-{x}{y}^{{{2}}}\right)}{\left.{d}{x}\right.}+{\left({x}+{x}^{{{2}}}{y}^{{{2}}}\right)}{\left.{d}{x}\right.}={0}\)
\(\displaystyle{\left({4}{\left.{d}{x}\right.}+{x}{\left.{d}{y}\right.}\right)}-{\left({x}{y}^{{{2}}}{\left.{d}{y}\right.}-{x}^{{{2}}}{y}^{{{2}}}\right)}={0}\)
\(\displaystyle{d}{\left({x}{y}\right)}-{\left({x}{y}\right)}^{{{2}}}{\left[{\frac{{{\left.{d}{x}\right.}}}{{{x}}}}-{\left.{d}{y}\right.}\right]}={0}\)
\(\displaystyle\int{\frac{{{d}{\left({x}{y}\right)}}}{{{\left({x}{y}\right)}^{{{2}}}}}}-\int{\frac{{{\left.{d}{x}\right.}}}{{{x}}}}+\int{\left.{d}{y}\right.}={0}\)
\(\displaystyle\Rightarrow-{\frac{{{1}}}{{{x}\cdot{y}}}}-{\ln{{x}}}+{y}={C}\)
put \(\displaystyle{4}{\left({1}\right)}={1}\)
\(\displaystyle-{\frac{{{1}}}{{{1}\cdot{1}}}}-{\ln{{1}}}+{1}={C}\Rightarrow-{1}-{0}+{1}={C}\)
\(\displaystyle{c}={0}\)
Hence, \(\displaystyle-{\frac{{{1}}}{{{x}\cdot{y}}}}\cdot{\ln{{x}}}+{y}={0}\)
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SlabydouluS62
Answered 2022-01-08 Author has 1786 answers
Consider the differential equation
\(\displaystyle{\left({y}-{x}{y}^{{{2}}}\right)}{\left.{d}{x}\right.}+{\left({x}+{x}^{{{2}}}{y}^{{{2}}}\right)}{\left.{d}{y}\right.}={0}\)
\(\displaystyle{y}{\left.{d}{x}\right.}-{x}{y}^{{{2}}}{\left.{d}{x}\right.}+{x}{\left.{d}{y}\right.}+{x}^{{{2}}}{y}^{{{2}}}{\left.{d}{y}\right.}={0}\)
\(\displaystyle{\left({y}{\left.{d}{x}\right.}+{x}{\left.{d}{y}\right.}\right)}+{\left({x}^{{{2}}}{y}^{{{2}}}{\left.{d}{y}\right.}-{x}{y}^{{{2}}}{\left.{d}{x}\right.}\right)}={0}\)
\(\displaystyle{d}{\left({x}{y}\right)}+{x}{y}{\left({x}{y}{\left.{d}{y}\right.}-{y}{\left.{d}{x}\right.}\right)}={0}\)
\(\displaystyle{d}{\left({x}{y}\right)}+{x}^{{{2}}}{y}^{{{2}}}{\left.{d}{y}\right.}-{x}{y}^{{{2}}}{\left.{d}{x}\right.}={0}\)
Dividing by \(\displaystyle{x}^{{{2}}}{y}^{{{2}}}\)
\(\displaystyle{d}\frac{{{x}{y}}}{{{x}^{{{2}}}{y}^{{{2}}}}}+{\left.{d}{y}\right.}-\frac{{\left.{d}{x}\right.}}{{x}}={0}\)
\(\displaystyle{d}\frac{{{x}{y}}}{{{\left({x}{y}\right)}^{{{2}}}}}+{\left.{d}{y}\right.}-\frac{{\left.{d}{x}\right.}}{{x}}={0}\)
Integrating the equation above we get
\(\displaystyle∫{\left[{d}\frac{{{x}{y}}}{{{\left({x}{y}\right)}^{{{2}}}}}\right]}+\int{\left.{d}{y}\right.}-\int\frac{{\left.{d}{x}\right.}}{{x}}={C}\)…..(1)
Where 'C' is constant of integration
Now say \(\displaystyle\int\frac{{\left.{d}{z}\right.}}{{z}^{{{2}}}}=-\frac{{1}}{{z}}\)
\(\displaystyle\int{x}^{{{n}}}{\left.{d}{x}\right.}={x}^{{{n}}}+\frac{{1}}{{{n}+{1}}}+{c}\ \text{for}\ {n}\ne-{1}\)
\(\displaystyle\int\frac{{\left.{d}{x}\right.}}{{x}}={\ln{{x}}}+{c}\)
Using these identities if \(\displaystyle{z}={x}{y}\) then the above eqn.(1) becomes
\(\displaystyle-\frac{{1}}{{{x}{y}}}+{y}-{\ln{{x}}}={C}\)
Multiply throughout by xy in the above equation we get
\(\displaystyle-{1}+{x}{y}^{{{2}}}-{x}{y}{\ln{{x}}}={C}{x}{y}\)
Therefore the solution of the equation is
\(\displaystyle{x}{y}^{{{2}}}-{x}{y}{\ln{{x}}}-{1}={C}{x}{y}\).
Where 'C' is any arbitrary constant.
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