ONE MORE STEP TOWARD BETTER TOMORROW : CHAPTER 12 - Examples on First Order Linear DEs

Tuesday, 5 August 2014

CHAPTER 12 - Examples on First Order Linear DEs

Example: 1

Solve the DE $\dfrac{{dx}}{{dy}} + \dfrac{x}{y} = {y^2}$

Solution: 1

Step-1

The I.F. is

	$I.F. = {e^{\int {P(y)dy} }}$	( $y$ is the ‘independent’ variable in this DE)
	$= {e^{\int {\dfrac{1}{y}dy} }}$
	$= {e^{\ln y}}$
	$= y$

Step-2

Multiplying by the I.F. on both sides, we have

	$y\dfrac{{dx}}{{dy}} + x = {y^3}$
	$\Rightarrow \dfrac{d}{{dy}}(xy) = {y^3}$
	$\Rightarrow d(xy) = {y^3}dy$

Step-3

Integrating both sides gives

$xy = \dfrac{{{y^4}}}{4} + C$

Example: 2

Solve the DE $\dfrac{{dy}}{{dx}} = {x^3}{y^3} - xy$

Solution: 2

Step-1

We have,

$\dfrac{{dy}}{{dx}} + xy = {x^3}{y^3}$

Note that since the RHS contains the term $y^3$ this DE is not in the standard linear DE form. However, a little artifice can enable us to reduce this to the standard form.

Divide both sides of the equation by $y^3$ :

$\dfrac{1}{{{y^3}}}\dfrac{{dy}}{{dx}} + \dfrac{x}{{{y^2}}} = {x^3}$

$\ldots (1)$

Step-2

Substitute $\dfrac{1}{y^2}=v$ :

	$\Rightarrow \dfrac{{ - 2}}{{{y^3}}}\dfrac{{dy}}{{dx}} = \dfrac{{dv}}{{dx}}$
	$\Rightarrow \dfrac{1}{{{y^3}}}\dfrac{{dy}}{{dx}} = \dfrac{{ - 1}}{2}\dfrac{{dv}}{{dx}}$	$\ldots (2)$