Jason Farmer
2021-02-08
Answered

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

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unessodopunsep

Answered 2021-02-09
Author has **105** answers

Here the Differential equations is given by

$(x+1)\frac{dy}{dx}=x({y}^{2}+1)$

$\to \frac{dy}{{y}^{2}+1}=\frac{xdx}{x+1}$

$\to \int \frac{dy}{{y}^{2}+1}=\int \frac{xdx}{x+1}$

$\to {\mathrm{tan}}^{-1}(y)=\int \frac{x+1-1dx}{(x+1)+c}$

$\to {\mathrm{tan}}^{-1}(y)=\int \frac{(x+1-1)dx)}{x+1+c}$

$\to \hspace{0.25em}{\mathrm{tan}}^{-1}(y)=\int dx-\int \frac{dx}{(x+1)+c}$

$\to {\mathrm{tan}}^{-1}(y)=x-\mathrm{ln}\mid x+1\mid +c$

$\to y=tan(x-\mathrm{ln}\mid x+1\mid +c),$

where cc is a arbitrary constant.

where cc is a arbitrary constant.

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