ONE MORE STEP TOWARD BETTER TOMORROW : CHAPTER 6 - Examples on Variable Separable Form

Tuesday, 5 August 2014

CHAPTER 6 - Examples on Variable Separable Form

Sometimes, the DE might not be in the variable-separable (VS) form; however, some manipulations might be able to transform it to a VS form. Lets see how this can be done. Consider the DE

$\dfrac{{dy}}{{dx}} = \cos \left( {x + y} \right)$

This is obviously not in VS form. Observe what happens if we use the following substitution in this DE:

	$x + y = v$
	$\Rightarrow 1 + \dfrac{{dy}}{{dx}} = \dfrac{{dv}}{{dx}}$

Thus, the DE transforms to

	$\dfrac{{dv}}{{dx}} - 1 = \cos v$
	$\Rightarrow \dfrac{{dv}}{{dx}} = 1 + \cos v$
	$\Rightarrow \dfrac{{dv}}{{1 + \cos v}} = dx$

which is clearly a VS form. Integrating both sides, we obtain

	$\int {\dfrac{{dv}}{{1 + \cos v}}} = \int {dx}$
	$\Rightarrow \dfrac{1}{2}\int {{{\sec }^2}\dfrac{v}{2}dv} = \int {dx}$
	$\Rightarrow \tan \dfrac{v}{2} = x + C$
	$\Rightarrow \tan \left( {\dfrac{{x + y}}{2}} \right) = x + C$

This is the required general solution to the DE.

From this example, you might be able to infer that any DE of the form

$\dfrac{{dy}}{{dx}} = f\left( {ax + by + c} \right)$

is reducible to a VS form using the technique described. Let us confirm this explicitly. Substitute

	$ax + by + c = v$
	$\Rightarrow a + b\dfrac{{dy}}{{dx}} = \dfrac{{dv}}{{dx}}$
	$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{b}\left( {\dfrac{{dv}}{{dx}} - a} \right)$

Thus, our DE reduces to

	$\dfrac{1}{b}\left( {\dfrac{{dv}}{{dx}} - a} \right) = f(v)$
	$\Rightarrow \dfrac{{dv}}{{dx}} = a + bf(v)$
	$\Rightarrow \dfrac{{dv}}{{a + bf(v)}} = dx$

which is obviously in VS form, and hence can be solved.

Example: 1

Solve the DE $\dfrac{{dy}}{{dx}} = \dfrac{{{r^2}}}{{{{\left( {x + y} \right)}^2}}}$

Solution: 1	Steps Involved: 2

Step-1

Substituting $x + y = v,$ we have

$\dfrac{{dy}}{{dx}} = \dfrac{{dv}}{{dx}} - 1$

and thus the DE reduces to

	$\dfrac{{dv}}{{dx}} - 1 = \dfrac{{{r^2}}}{{{v^2}}}$
	$\Rightarrow \dfrac{{{v^2}}}{{{r^2} + {v^2}}}dv = dx$
	$\Rightarrow \left( {1 - \dfrac{{{r^2}}}{{{r^2} + {v^2}}}} \right)dv = dx$

Step-2

Integrating, we have

	$v - r{\tan ^{ - 1}}\left( {\dfrac{v}{r}} \right) = x + C$
	$\Rightarrow (x + y) - r{\tan ^{ - 1}}\left( {\dfrac{{x + y}}{r}} \right) = x + C$

Example: 2

Solve the DE $\dfrac{{dy}}{{dx}} = \dfrac{{\left( {x + y} \right) + \left( {x + y - 1} \right)\ln \left( {x + y} \right)}}{{\ln \left( {x + y} \right)}}$

Solution: 2	Steps Involved: 3

Step-1

Again, the substitution $x + y = v$ will reduce this DE to the following VS form:

	$\dfrac{{dv}}{{dx}} - 1 = \dfrac{{v + \left( {v - 1} \right)\ln v}}{{\ln v}}$
	$= \left( {v - 1} \right) + \dfrac{v}{{\ln v}}$
	$\Rightarrow \dfrac{{dv}}{{dx}} = v + \dfrac{v}{{\ln v}}$
	$\Rightarrow \dfrac{{\ln v}}{{v\left( {1 + \ln v} \right)}}dv = dx$

Step-2

Integrating, we have

$\int {\dfrac{{\ln v}}{{v\left( {1 + \ln v} \right)}}dv} = \int {dx}$

Step-3

To evaluate the integral on the LHS, we use the substitution $\left( {1 + \ln v} \right) = t$ which gives $\dfrac{1}{v}dv = dt$ . Thus,

	${\dfrac{{t - 1}}{t}dt = \int {dx} }$
	$\Rightarrow t - \ln t = x + C$
	$\Rightarrow \left( {1 + \ln v} \right) - \ln \left( {1 + \ln v} \right) = x + C$
	$\Rightarrow \left( {1 + \ln \left( {x + y} \right)} \right) - \ln \left( {1 + \ln \left( {x + y} \right)} \right) = x + C$

Tuesday, 5 August 2014

CHAPTER 6 - Examples on Variable Separable Form

Step-1

Step-2

Step-1

Step-2

Step-3

No comments:

https://www.youtube.com/TarunGehlot