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

Question
Differential equations
Q. 2# $$\displaystyle{\left({x}+{1}\right)}{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={x}{\left({y}^{{{2}}}+{1}\right)}$$

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{ ### Relevant Questions asked 2021-03-22 Solve the equation: \(\displaystyle{\left({x}+{1}\right)}{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={x}{\left({y}^{{2}}+{1}\right)}$$
Solve the equation:
$$\displaystyle{\left({a}-{x}\right)}{\left.{d}{y}\right.}+{\left({a}+{y}\right)}{\left.{d}{x}\right.}={0}$$
Q. 2# $$(x+1)\frac{dy}{dx}=x(y^{2}+1)$$
Calculating derivatives Find dy>dx for the following functions.
$$\displaystyle{y}={\frac{{{1}}}{{{2}+{\sin{{x}}}}}}$$
The graph of y = f(x) contains the point (0,2), $$\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={\frac{{-{x}}}{{{y}{e}^{{{x}^{{2}}}}}}}$$, and f(x) is greater than 0 for all x, then f(x)=
A) $$\displaystyle{3}+{e}^{{-{x}^{{2}}}}$$
B) $$\displaystyle\sqrt{{{3}}}+{e}^{{-{x}}}$$
C) $$\displaystyle{1}+{e}^{{-{x}}}$$
D) $$\displaystyle\sqrt{{{3}+{e}^{{-{x}^{{2}}}}}}$$
E) $$\displaystyle\sqrt{{{3}+{e}^{{{x}^{{2}}}}}}$$
Find the differential dy for the given values of x and dx. $$y=\frac{e^x}{10},x=0,dx=0.1$$
Q. 1# $$(a−x)dy+(a+y)dx=0(a−x)dy+(a+y)dx=0$$
Q. 1# $$\displaystyle{\left({a}−{x}\right)}{\left.{d}{y}\right.}+{\left({a}+{y}\right)}{\left.{d}{x}\right.}={0}{\left({a}−{x}\right)}{\left.{d}{y}\right.}+{\left({a}+{y}\right)}{\left.{d}{x}\right.}={0}$$
Find $$\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}$$ using the rules for finding derivatives.
$$\displaystyle{y}={\left({2}{x}^{{{2}}}-{6}\right)}{\left({8}{x}^{{{2}}}-{9}{x}+{9}\right)}$$
$$\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}=$$
Make and solve the given equation $$x\ dx\ +\ y\ dy=a^{2}\frac{x\ dy\ -\ y\ dx}{x^{2}\ +\ y^{2}}$$