tgt

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