# Examples On Differential Equations In Variable Separable Form

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Example -5

Solve the DE\begin{align}x{y^2}\frac{{dy}}{{dx}} = 1 - {x^2} + {y^2} - {x^2}{y^2}.\end{align}

Solution: Again, this DE is of the variable separable form as can be made evident by a slight rearrangement.

\begin{align} &\qquad \;\;\;\;x{y^2}\frac{{dy}}{{dx}} = (1 - {x^2})(1 + {y^2})\\&\Rightarrow \quad \left( {\frac{{{y^2}}}{{1 + {y^2}}}} \right)dy = \left( {\frac{{1 - {x^2}}}{x}} \right)dx\\& \Rightarrow \quad \left( {1 - \frac{1}{{1 + {y^2}}}} \right)dy = \left( {\frac{1}{x} - x} \right)dx\end{align}

Integrating both sides, we have

$y - {\tan ^{ - 1}}y = \ln x - \frac{{{x^2}}}{2} + C$

This is the required general solution.

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

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

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

\begin{align} &\qquad \;\;x + y = v\\&\Rightarrow \quad 1 + \frac{{dy}}{{dx}} = \frac{{dv}}{{dx}}\end{align}

Thus, the DE transforms to

\begin{align} &\qquad\;\; \frac{{dv}}{{dx}} - 1 = \cos v\\& \Rightarrow \quad \frac{{dv}}{{dx}} = 1 + \cos v\\ &\Rightarrow \quad \frac{{dv}}{{1 + \cos v}} = dx\end{align}

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

\begin{align} &\qquad \;\;\int {\frac{{dv}}{{1 + \cos v}}} = \int {dx} \\&\Rightarrow \quad \frac{1}{2}\int {{{\sec }^2}\frac{v}{2}dv} = \int {dx} \\& \Rightarrow \quad \tan \frac{v}{2} = x + C\\ &\Rightarrow \quad \tan \left( {\frac{{x + y}}{2}} \right) = x + C\end{align}

This is the required general solution to the DE.

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

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

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

Substitute

\begin{align} &\qquad \;\;ax + by + c = v\\ &\Rightarrow \quad a + b\frac{{dy}}{{dx}} = \frac{{dv}}{{dx}}\\ &\Rightarrow \quad \frac{{dy}}{{dx}} = \frac{1}{b}\left( {\frac{{dv}}{{dx}} - a} \right)\end{align}

Thus, our DE reduces to

\begin{align}&\qquad \frac{1}{b}\left( {\frac{{dv}}{{dx}} - a} \right) = f(v)\\& \Rightarrow \quad \frac{{dv}}{{dx}} = a + bf(v)\\&\Rightarrow \quad \frac{{dv}}{{a + bf(v)}} = dx\end{align}

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

Example - 6

Solve the DE \begin{align}\frac{{dy}}{{dx}} = \frac{{{r^2}}}{{{{\left( {x + y} \right)}^2}}}.\end{align}

Solution: Substituting $$x + y = v,$$ we have

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

and thus the DE reduces to

\begin{align} &\qquad\;\; \frac{{dv}}{{dx}} - 1 = \frac{{{r^2}}}{{{v^2}}}\\& \Rightarrow \quad \frac{{{v^2}}}{{{r^2} + {v^2}}}dv = dx\\ &\Rightarrow \quad \left( {1 - \frac{{{r^2}}}{{{r^2} + {v^2}}}} \right)dv = dx\end{align}

Integrating, we have

\begin{align} &\qquad \;\;v - r{\tan ^{ - 1}}\left( {\frac{v}{r}} \right) = x + C\\&\Rightarrow \quad (x + y) - r{\tan ^{ - 1}}\left( {\frac{{x + y}}{r}} \right) = x + C\end{align}

Example - 7

Solve the DE \begin{align}\frac{{dy}}{{dx}} = \frac{{\left( {x + y} \right) + \left( {x + y - 1} \right)\ln \left( {x + y} \right)}}{{\ln \left( {x + y} \right)}}\end{align}.

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

\begin{align}&\qquad \;\frac{{dv}}{{dx}} - 1 = \frac{{v + \left( {v - 1} \right)\ln v}}{{\ln v}}\\ &\quad\qquad\qquad\;\, = \left( {v - 1} \right) + \frac{v}{{\ln v}}\\&\Rightarrow \quad\frac{{dv}}{{dx}} = v + \frac{v}{{\ln v}}\\&\Rightarrow \quad \frac{{\ln v}}{{v\left( {1 + \ln v} \right)}}dv = dx\end{align}

Integrating, we have

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

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

\begin{align} &\qquad \; \int {\frac{{t - 1}}{t}dt = \int {dx} } \\&\Rightarrow \quad t - \ln t = x + C\\&\Rightarrow \quad \left( {1 + \ln v} \right) - \ln \left( {1 + \ln v} \right) = x + C\\&\Rightarrow \quad \left( {1 + \ln \left( {x + y} \right)} \right) - \ln \left( {1 + \ln \left( {x + y} \right)} \right) = x + C\end{align}

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