Notes: `xy + y' = 100x` Solve the differential equation

Monday, 12 June 2017

$\displaystyle{x}{y}+{y}'={100}{x}$ Solve the differential equation

The problem: $\displaystyle{x}{y}+{y}'={100}{x}$ is as first order differential equation that we can evaluate by applying variable separable differential equation:

$\displaystyle{N}{\left({y}\right)}{y}'={M}{\left({x}\right)}$

$\displaystyle{N}{\left({y}\right)}\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}={M}{\left({x}\right)}$

$\displaystyle{N}{\left({y}\right)}{\left.{d}{y}\right.}={M}{\left({x}\right)}{\left.{d}{x}\right.}$

Apply direct integration: $\displaystyle\int{N}{\left({y}\right)}{\left.{d}{y}\right.}=\int{M}{\left({x}\right)}{\left.{d}{x}\right.}$ to solve for the

general solution of a differential equation.

Applying variable separable differential equation, we get:

$\displaystyle{x}{y}+{y}'={100}{x}$

$\displaystyle{y}'={100}{x}-{x}{y}$

$\displaystyle{y}'={x}{\left({100}-{y}\right)}$

$\displaystyle\frac{{{y}'}}{{{100}-{y}}}={x}$

Let $\displaystyle{y}'=\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}$ :

$\displaystyle\frac{{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}}{{{100}-{y}}}={x}$

$\displaystyle\frac{{{\left.{d}{y}\right.}}}{{{100}-{y}}}={x}{\left.{d}{x}\right.}$

Apply direct integration on both sides:

$\displaystyle\int\frac{{{\left.{d}{y}\right.}}}{{{100}-{y}}}=\int{x}{\left.{d}{x}\right.}$

...

The problem: $\displaystyle{x}{y}+{y}'={100}{x}$ is as first order differential equation that we can evaluate by applying variable separable differential equation:

$\displaystyle{N}{\left({y}\right)}{y}'={M}{\left({x}\right)}$

$\displaystyle{N}{\left({y}\right)}\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}={M}{\left({x}\right)}$

$\displaystyle{N}{\left({y}\right)}{\left.{d}{y}\right.}={M}{\left({x}\right)}{\left.{d}{x}\right.}$

Apply direct integration: $\displaystyle\int{N}{\left({y}\right)}{\left.{d}{y}\right.}=\int{M}{\left({x}\right)}{\left.{d}{x}\right.}$ to solve for the

general solution of a differential equation.

Applying variable separable differential equation, we get:

$\displaystyle{x}{y}+{y}'={100}{x}$

$\displaystyle{y}'={100}{x}-{x}{y}$

$\displaystyle{y}'={x}{\left({100}-{y}\right)}$

$\displaystyle\frac{{{y}'}}{{{100}-{y}}}={x}$

Let $\displaystyle{y}'=\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}$ :

$\displaystyle\frac{{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}}{{{100}-{y}}}={x}$

$\displaystyle\frac{{{\left.{d}{y}\right.}}}{{{100}-{y}}}={x}{\left.{d}{x}\right.}$

Apply direct integration on both sides:

$\displaystyle\int\frac{{{\left.{d}{y}\right.}}}{{{100}-{y}}}=\int{x}{\left.{d}{x}\right.}$

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

$\displaystyle{u}={100}-{y}$ then $\displaystyle{d}{u}=-{\left.{d}{y}\right.}$ or - $\displaystyle{d}{u}={\left.{d}{y}\right.}.$

The integral becomes:

$\displaystyle\int\frac{{{\left.{d}{y}\right.}}}{{{100}-{y}}}=\int\frac{{-{d}{u}}}{{{u}}}$

Applying basic integration formula for logarithm:

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

Plug-in $\displaystyle{u}={100}-{y}$ on " $\displaystyle-{\ln}{\left|{u}\right|}$ " , we get:

$\displaystyle\int\frac{{{\left.{d}{y}\right.}}}{{{100}-{y}}}=-{\ln}{\left|{100}-{y}\right|}$

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