Question

Find the first order differential eqn of y
dy/dx = y(x² + 1)​

Answer 1

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

- Simplify by collecting each term according to its corresponding d

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

- Integrate both sides;

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

- To make y the subject, you must remove the natural log;

`{ \tt{ log_(e)(y) = (1)/(3) {x}^(3) + x + c }} \\ \\ { \tt{y = {e}^{( (1)/(3) {x}^(3) + x + c) } }} \\`

Answer 2

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

Explanation:

Given differential equation:

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

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:

`\large\text{$\implies (1)/(y)\;\text{d}y=(x^2+1)\;\text{d}x$}`

Integrate both sides, remembering to add the constant of integration (C):

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

Rewrite C as ln k:

`\large\text{$\implies \ln y = (1)/(3)x^3+x+\ln k$}`

Solve for y, applying:

• 
  `\textsf{Log rule}: \quad e^(\ln a)=a`

• 
  `\textsf{Exponent rule}: \quad \:a^(b+c)=a^ba^c`

`\large\text{$ \implies e^(\ln y) =e^{(1)/(3)x^3+x+\ln k}$}`

`\large\text{$ \implies y=e^(\ln k) \cdot e^{(1)/(3)x^3+x}$}`

`\large\text{$\implies y=ke^{(1)/(3)x^3+x}$}`

Integration rules

`\boxed{\begin{minipage}{3.6 cm}\underline{Integrating $(1)/(x)$}\\\\$\displaystyle \int (1)/(x)\:\text{d}x=\ln |x|+\text{C}$\end{minipage}}`

`\boxed{\begin{minipage}{3.6 cm}\underline{Integrating $x^n$}\\\\$\displaystyle \int x^n\:\text{d}x=(x^(n+1))/(n+1)+\text{C}$\end{minipage}}`

`\boxed{\begin{minipage}{5.1 cm}\underline{Integrating a constant}\\\\$\displaystyle \int n\:\text{d}x=nx+\text{C}$\\\\(where $n$ is any constant value) \end{minipage}}`