Answer:
[tex]{ \sf{ \frac{dy}{dx} = y( {x}^{2} + 1)}} \\ \\ { \sf{ \frac{dy}{y} = ( {x}^{2} + 1)dx}} \\ \\ { \sf{ \int \frac{1}{y} dy = \int ( {x}^{2} + 1) \: dx}} \\ \\ { \sf{ ln(y) = \frac{ {x}^{3} }{3} + x + c }}[/tex]
Answer:
[tex]\large\text{$y& = e^{\frac{1}{3}x^3+x+\text{C}}$}[/tex]
Step-by-step explanation:
Given differential equation:
[tex]\dfrac{\text{d}x}{\text{d}y}=\dfrac{1}{y(x^2+1)}[/tex]
Rearrange the equation so that all the terms containing y are on the left side, and all the terms containing x are on the right side:
[tex]\begin{aligned}\implies \dfrac{\text{d}x}{\text{d}y}&=\dfrac{1}{y(x^2+1)}\\\\(x^2+1)\;\dfrac{\text{d}x}{\text{d}y}&=\dfrac{1}{y}\\\\(x^2+1)\;\text{d}x}&=\dfrac{1}{y}\;\text{d}y\\\\\dfrac{1}{y}\;\text{d}y&=(x^2+1)\;\text{d}x}\end{aligned}[/tex]
Integrate both sides:
[tex]\begin{aligned}\implies \displaystyle \int \dfrac{1}{y}\;\text{d}y &= \int (x^2+1)\;\text{d}x}\\\\\int \dfrac{1}{y}\;\text{d}y &= \int x^2\;\text{d}x}+\int 1\;\text{d}x}\\\\\ln y & = \dfrac{1}{3}x^3+x+\text{C}\\\\e^{\ln y} & = e^{\frac{1}{3}x^3+x+\text{C}}\\\\y& = e^{\frac{1}{3}x^3+x+\text{C}}\\\\\end{aligned}[/tex]
Therefore, the solution to the given differential equation is:
[tex]\large\text{$y& = e^{\frac{1}{3}x^3+x+\text{C}}$}[/tex]
Integration rules used:
[tex]\boxed{\begin{minipage}{3.5 cm}\underline{Integrating $\frac{1}{x}$}\\\\$\displaystyle \int \dfrac{1}{x}\:\text{d}x=\ln |x|+\text{C}$\end{minipage}}[/tex]
[tex]\boxed{\begin{minipage}{3.5 cm}\underline{Integrating $x^n$}\\\\$\displaystyle \int x^n\:\text{d}x=\dfrac{x^{n+1}}{n+1}+\text{C}$\end{minipage}}[/tex]
[tex]\boxed{\begin{minipage}{5 cm}\underline{Integrating a constant}\\\\$\displaystyle \int n\:\text{d}x=nx+\text{C}$\\(where $n$ is any constant value)\end{minipage}}[/tex]
