Question

Solve the differential equation


`(dx)/(dy) = \frac{1}{y( {x}^(2) + 1)} \\`
​

Answer 1

`{ \sf{ (dy)/(dx) = y( {x}^(2) + 1)}} \\ \\ { \sf{ (dy)/(y) = ( {x}^(2) + 1)dx}} \\ \\ { \sf{ \int (1)/(y) dy = \int ( {x}^(2) + 1) \: dx}} \\ \\ { \sf{ ln(y) = \frac{ {x}^(3) }{3} + x + c }}`

Answer 2

`\large\text{$y& = e^{(1)/(3)x^3+x+\text{C}}$}`

Explanation:

Given differential equation:

`\frac{\text{d}x}{\text{d}y}=(1)/(y(x^2+1))`

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:

`\begin{aligned}\implies \frac{\text{d}x}{\text{d}y}&=(1)/(y(x^2+1))\\\\(x^2+1)\;\frac{\text{d}x}{\text{d}y}&=(1)/(y)\\\\(x^2+1)\;\text{d}x}&=(1)/(y)\;\text{d}y\\\\(1)/(y)\;\text{d}y&=(x^2+1)\;\text{d}x}\end{aligned}`

Integrate both sides:

`\begin{aligned}\implies \displaystyle \int (1)/(y)\;\text{d}y &= \int (x^2+1)\;\text{d}x}\\\\\int (1)/(y)\;\text{d}y &= \int x^2\;\text{d}x}+\int 1\;\text{d}x}\\\\\ln y & = (1)/(3)x^3+x+\text{C}\\\\e^(\ln y) & = e^{(1)/(3)x^3+x+\text{C}}\\\\y& = e^{(1)/(3)x^3+x+\text{C}}\\\\\end{aligned}`

Therefore, the solution to the given differential equation is:

`\large\text{$y& = e^{(1)/(3)x^3+x+\text{C}}$}`

Integration rules used:

`\boxed{\begin{minipage}{3.5 cm}\underline{Integrating $(1)/(x)$}\\\\$\displaystyle \int (1)/(x)\:\text{d}x=\ln |x|+\text{C}$\end{minipage}}`

`\boxed{\begin{minipage}{3.5 cm}\underline{Integrating $x^n$}\\\\$\displaystyle \int x^n\:\text{d}x=(x^(n+1))/(n+1)+\text{C}$\end{minipage}}`

`\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}}`