Q. 2# (x+1)frac{dy}{dx}=x(y^{2}+1)

Q. 2# (x+1)frac{dy}{dx}=x(y^{2}+1)

Question
Differential equations
asked 2020-12-25
Q. 2# \(\displaystyle{\left({x}+{1}\right)}{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={x}{\left({y}^{{{2}}}+{1}\right)}\)

Answers (1)

2020-12-26
Here the Differential equations is given by
\(\displaystyle{\left({x}+{1}\right)}{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={x}{\left({y}^{{{2}}}+{1}\right)}\)
\(\displaystyle\rightarrow{\frac{{{\left.{d}{y}\right.}}}{{{y}^{{{2}}}+{1}}}}={\frac{{{x}{\left.{d}{x}\right.}}}{{{x}+{1}}}}\)
\(\displaystyle\rightarrow\int{\frac{{{\left.{d}{y}\right.}}}{{{y}^{{{2}}}+{1}}}}=\int{\frac{{{x}{\left.{d}{x}\right.}}}{{{x}+{1}}}}\)
\(\displaystyle\rightarrow{{\tan}^{{−{1}}}{\left({y}\right)}}=\int{\frac{{{x}+{1}−{1}{\left.{d}{x}\right.}}}{{{\left({x}+{1}\right)}+{c}}}}\)
\(\displaystyle\rightarrow{{\tan}^{{−{1}}}{\left({y}\right)}}=\int{\frac{{{\left({x}+{1}−{1}\right)}{\left.{d}{x}\right.}}}{\rbrace}}{\left\lbrace{x}+{1}+{c}\right\rbrace}\)
\(\displaystyle\rightarrow {{\tan}^{{−{1}}}{\left({y}\right)}}=\int{\left.{d}{x}\right.}−\int{\frac{{{\left.{d}{x}\right.}}}{{{\left({x}+{1}\right)}+{c}}}}\)
\(\displaystyle\rightarrow{{\tan}^{{−{1}}}{\left({y}\right)}}={x}−{\ln{
0

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