Notes: `(2+x)y' = 3y` Find the general solution of the differential equation

Tuesday, 13 May 2014

$\displaystyle{\left({2}+{x}\right)}{y}'={3}{y}$ Find the general solution of the differential equation

Recall that $\displaystyle{y}'$ is the same as $\displaystyle\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}$ . Then in the given problem: $\displaystyle{\left({2}+{x}\right)}{y}'={3}{y}$ , we may write it as:

$\displaystyle{\left({2}+{x}\right)}\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}={3}{y}.$

This will help to follow the variable separable differential equation in a form of $\displaystyle{N}{\left({y}\right)}{\left.{d}{y}\right.}={M}{\left({x}\right)}{\left.{d}{x}\right.}.$

To rearrange $\displaystyle{\left({2}+{x}\right)}\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}={3}{y}$ ,cross-multiply $\displaystyle{\left({\left.{d}{x}\right.}\right)}$ to the other side:

$\displaystyle{\left({2}+{x}\right)}{\left.{d}{y}\right.}={3}{y}{\left.{d}{x}\right.}$

Divide both sides by $\displaystyle{\left({2}+{x}\right)}$ :

$\displaystyle\frac{{{\left({2}+{x}\right)}{\left.{d}{y}\right.}}}{{{2}+{x}}}=\frac{{{3}{y}{\left.{d}{x}\right.}}}{{{2}+{x}}}$

$\displaystyle{\left.{d}{y}\right.}=\frac{{{3}{y}{\left.{d}{x}\right.}}}{{{2}+{x}}}$

Divide both sides by $\displaystyle{y}$ :

...

$\displaystyle{\left({2}+{x}\right)}\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}={3}{y}.$

This will help to follow the variable separable differential equation in a form of $\displaystyle{N}{\left({y}\right)}{\left.{d}{y}\right.}={M}{\left({x}\right)}{\left.{d}{x}\right.}.$

To rearrange $\displaystyle{\left({2}+{x}\right)}\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}={3}{y}$ ,cross-multiply $\displaystyle{\left({\left.{d}{x}\right.}\right)}$ to the other side:

$\displaystyle{\left({2}+{x}\right)}{\left.{d}{y}\right.}={3}{y}{\left.{d}{x}\right.}$

Divide both sides by $\displaystyle{\left({2}+{x}\right)}$ :

$\displaystyle\frac{{{\left({2}+{x}\right)}{\left.{d}{y}\right.}}}{{{2}+{x}}}=\frac{{{3}{y}{\left.{d}{x}\right.}}}{{{2}+{x}}}$

$\displaystyle{\left.{d}{y}\right.}=\frac{{{3}{y}{\left.{d}{x}\right.}}}{{{2}+{x}}}$

Divide both sides by $\displaystyle{y}$ :

$\displaystyle\frac{{{\left.{d}{y}\right.}}}{{y}}=\frac{{{3}{y}{\left.{d}{x}\right.}}}{{{\left({2}+{x}\right)}{y}}}$

$\displaystyle\frac{{{\left.{d}{y}\right.}}}{{y}}=\frac{{{3}{\left.{d}{x}\right.}}}{{{2}+{x}}}$

To solve for the general solution of the differential equation, apply direct integration on both sides:

$\displaystyle\int\frac{{{\left.{d}{y}\right.}}}{{y}}=\int\frac{{{3}{\left.{d}{x}\right.}}}{{{2}+{x}}}$

For the left side, apply the basic integration formula for logarithm:

$\displaystyle\int\frac{{{\left.{d}{y}\right.}}}{{y}}={\ln}{\left|{y}\right|}$

For the right side, we may apply the basic integration property: $\displaystyle\int{c}\cdot{f{{\left({x}\right)}}}{\left.{d}{x}\right.}={c}\int{f{{\left({x}\right)}}}{\left.{d}{x}\right.}$ .

$\displaystyle\int\frac{{{3}{\left.{d}{x}\right.}}}{{{2}+{x}}}={3}\int\frac{{{\left.{d}{x}\right.}}}{{{2}+{x}}}$

Let $\displaystyle{u}={2}+{x}$ then du= dx

The integral becomes:

$\displaystyle{3}\int\frac{{{\left.{d}{x}\right.}}}{{{2}+{x}}}={3}\int\frac{{{d}{u}}}{{u}}$

We can now apply the basic integration formula for logarithm on the integral part:

$\displaystyle{3}\int\frac{{{d}{u}}}{{u}}={3}{\ln}{\left|{u}\right|}+{C}$

Recall $\displaystyle{u}={\left({2}+{x}\right)}$ then $\displaystyle{3}\int\frac{{{\left.{d}{x}\right.}}}{{{2}+{x}}}={3}{\ln}{\left|{2}+{x}\right|}+{C}$