`y' = x(1+y)` Solve the differential equation

Wednesday, March 5, 2014

$\displaystyle{y}'={x}{\left({1}+{y}\right)}$ Solve the differential equation

An ordinary differential equation (ODE) has differential equation for a function with single variable. A first order ODE follows $\displaystyle\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}={f{{\left({x},{y}\right)}}}$ .

It can also be in a form of $\displaystyle{N}{\left({y}\right)}{\left.{d}{y}\right.}={M}{\left({x}\right)}{\left.{d}{x}\right.}$ as variable separable differential equation.

To be able to set-up the problem as $\displaystyle{N}{\left({y}\right)}{\left.{d}{y}\right.}={M}{\left({x}\right)}{\left.{d}{x}\right.}$ , we let $\displaystyle{y}'=\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}$ .

The problem: $\displaystyle{y}'={x}{\left({1}+{y}\right)}$ becomes:

$\displaystyle\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}={x}{\left({1}+{y}\right)}$

Rearrange by cross-multiplication, we get:

$\displaystyle\frac{{{\left.{d}{y}\right.}}}{{{1}+{y}}}={x}{\left.{d}{x}\right.}$

Apply direct integration on both sides: $\displaystyle\int\frac{{{\left.{d}{y}\right.}}}{{{1}+{y}}}=\int{x}{\left.{d}{x}\right.}$ to solve for the general solution of a differential equation.

For the left side, we consider u-substitution by letting:

$\displaystyle{u}={1}+{y}$ then $\displaystyle{d}{u}={\left.{d}{y}\right.}$

The integral becomes: $\displaystyle\int\frac{{{\left.{d}{y}\right.}}}{{{1}+{y}}}=\int\frac{{{d}{u}}}{{{u}}}$

Applying basic integration formula for logarithm:

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

Plug-in $\displaystyle{u}={1}+{y}$ on $\displaystyle{\ln}{\left|{u}\right|}$ , we get:

$\displaystyle\int\frac{{{\left.{d}{y}\right.}}}{{{1}+{y}}}={\ln}{\left|{1}+{y}\right|}$

For the right side, we apply the Power Rule of integration: $\displaystyle\int{{x}}^{{n}}{\left.{d}{x}\right.}=\frac{{{x}}^{{{n}+{1}}}}{{{n}+{1}}}+{C}$