Solve differential equation \frac{dy}{dx}= (x+y+1)^2-(x+y-1)^2

Solve differential equation \frac{dy}{dx}= (x+y+1)^2-(x+y-1)^2

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

Answers (1)

2021-03-08
On substituting
\(\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={y}^{{t}}\)
\(\displaystyle{y}^{{t}}={\left({x}+{y}+{1}\right)}^{{2}}-{\left({x}+{y}-{1}\right)}^{{2}}\)
\(\displaystyle{y}^{{t}}-{4}{y}={4}{x}\)
On finding the integration factor
\(\displaystyle\mu{\left({x}\right)}={e}^{{-{{4}}}}{x}\)
Thus on writing the equation in the form
\(\displaystyle{\left(\mu{\left({x}\right)}{y}\right)}'=\mu{\left({x}\right)}{q}{\left({x}\right)}\)
\(\displaystyle{\left({e}^{{-{4}}}{y}\right)}'={4}{x}{e}^{{-{4}}}\)
\(\displaystyle{y}=-{x}-{\frac{{{1}}}{{{4}}}}+{c}_{{1}}{e}^{{{4}{x}}}\)
0

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