Solve differential equation 2xy-9x^2+(2y+x^2+1)\frac{dy}{dx}=0, \ y(0)=-3

Solve differential equation 2xy-9x^2+(2y+x^2+1)\frac{dy}{dx}=0, \ y(0)=-3

Question
Solve differential equation \(\displaystyle{2}{x}{y}-{9}{x}^{{2}}+{\left({2}{y}+{x}^{{2}}+{1}\right)}{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={0},\ {y}{\left({0}\right)}=-{3}\)

Answers (1)

2020-12-25
The equation cam be written as
\(\displaystyle{\left({2}{x}{y}-{9}{x}^{{2}}\right)}{\left.{d}{x}\right.}+{\left({2}{y}+{x}^{{2}}+{1}\right)}{\left.{d}{y}\right.}={0}\)
The equation is of the form
\(\displaystyle{M}{\left.{d}{x}\right.}+{N}{\left.{d}{y}\right.}={0}\)
\(\displaystyle{M}{y}={2}{x}\)
\(\displaystyle{N}{x}={2}{x}\)
The equation is exact since My=Nx
Assume that F(x,y) is the solution for the equation
\(\displaystyle{F}{x}={M}\)
\(\displaystyle={2}{x}{y}-{9}{x}^{{2}}\)
\(\displaystyle{F}=\int{2}{x}{y}-{9}{x}^{{2}}{\left.{d}{x}\right.}\)
\(\displaystyle{F}={x}^{{2}}{y}-{3}{x}^{{2}}+{g{{\left({y}\right)}}}\)
Take the derivative for F with respect to y
\(\displaystyle{F}{y}={x}^{{2}}+{g}'{\left({y}\right)}\)
\(\displaystyle{2}{y}+{x}^{{2}}+{1}={g}'{\left({y}\right)}\)
\(\displaystyle{g}'{\left({y}\right)}={2}{y}+{1}\)
\(\displaystyle{g}'{\left({y}\right)}=\int{2}{y}+{1}{\left.{d}{y}\right.}\)
\(\displaystyle{g}'{\left({y}\right)}={y}^{{2}}+{y}\)
Hence
\(\displaystyle{x}^{{2}}{y}-{3}{x}^{{3}}+{y}^{{2}}+{y}={C}\)
\(\displaystyle{y}^{{2}}+{\left({x}^{{2}}-{1}\right)}{y}-{3}{x}^{{2}}={C}\)
It is known that
\(\displaystyle{y}{\left({0}\right)}=-{3}\)
\(\displaystyle{9}-{3}={c}\)
\(\displaystyle{c}={6}\)
So the solution for the problem is
\(\displaystyle{y}^{{2}}{\left({x}^{{2}}-{1}\right)}{y}-{3}{x}^{{2}}-{6}={0}\)
0

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