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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: 1Steps 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: 2Steps 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) = twhich 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
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