Question

Find the particular solution of the differential equation
\[\left( {x - y} \right)\dfrac{{dy}}{{dx}} = \left( {x + 2y} \right)\]
given that \[y = 0\] when \[x = 1\]

Answer 1

Hint: First we need to identify the category of the differential equation. If we replace x by \[\lambda x\] and y by \[\lambda y\], we find that the differential equation doesn’t change.
Hence, it is a homogeneous differential equation. We can then use the standard method to solve this differential equation. Finding a particular solution implies that the final solution should be independent of any arbitrary constant.

Complete step-by-step solution:
Given: The differential equation is given as \[\left( {x - y} \right)\dfrac{{dy}}{{dx}} = \left( {x + 2y} \right)\] and \[y = 0\]when \[x = 1\].
We can write the differential equation as:
\[\dfrac{{dy}}{{dx}} = \dfrac{{x + 2y}}{{x - y}} .............\left( i \right)\]
Let us put\[y = \vartheta x .............\left( {ii} \right)\]
Differentiating both sides of (ii) with respect to x, we get
\[
    \dfrac{{dy}}{{dx}} = \dfrac{{d\left( {\vartheta x} \right)}}{{dx}} \\
   \Rightarrow \dfrac{{dy}}{{dx}} = \vartheta \dfrac{{dx}}{{dx}} + x\dfrac{{d\vartheta }}{{dx}} \\
   \Rightarrow \dfrac{{dy}}{{dx}} = \vartheta + x\dfrac{{d\vartheta }}{{dx}} ......\left( {ii} \right) \\
 \]
Putting the expression of y and \[\dfrac{{dy}}{{dx}}\] in, we get
\[
    \vartheta + x\dfrac{{d\vartheta }}{{dx}} = \dfrac{{x + 2\vartheta x}}{{x - \vartheta x}} \\
   \Rightarrow \vartheta + x\dfrac{{d\vartheta }}{{dx}} = \dfrac{{1 + 2\vartheta }}{{1 - \vartheta }} \\
   \Rightarrow x\dfrac{{d\vartheta }}{{dx}} = \dfrac{{1 + 2\vartheta }}{{1 - \vartheta }} - \vartheta \\
   \Rightarrow x\dfrac{{d\vartheta }}{{dx}} = \dfrac{{1 + 2\vartheta - \vartheta + {\vartheta ^2}}}{{1 - \vartheta }} \\
 \]
\[
   \Rightarrow x\dfrac{{d\vartheta }}{{dx}} = \dfrac{{{\vartheta ^2} + \vartheta + 1}}{{1 - \vartheta }} \\
   \Rightarrow \dfrac{{\left( {1 - \vartheta } \right)d\vartheta }}{{{\vartheta ^2} + \vartheta + 1}} = \dfrac{{dx}}{x} \\
 \]
Integrating both sides
\[
    \int {\dfrac{{\left( {1 - \vartheta } \right)d\vartheta }}{{{\vartheta ^2} + \vartheta + 1}}} = \int {\dfrac{{dx}}{x}} \\
    1 - \vartheta \, = P\dfrac{d}{{d\vartheta }}\left( {{\vartheta ^2} + \vartheta + 1} \right) + Q \\
   \Rightarrow - \vartheta + 1 = P\left( {2\vartheta + 1} \right) + Q \\
   \Rightarrow - \vartheta + 1 = 2P\vartheta + \left( {P + Q} \right) \\
 \]
On comparing both sides, we get
\[
   \Rightarrow 2P = - 1 \\
   \Rightarrow P = - \dfrac{1}{2} \\
    P + Q = 1 \\
   \Rightarrow - \dfrac{1}{2} + Q = 1 \\
   \Rightarrow Q = \dfrac{3}{2} \\
  \therefore 1 - \vartheta = - \dfrac{1}{2}\left( {2\vartheta + 1} \right) + \dfrac{3}{2} \\
 \]
Thus,
\[
    \int {\dfrac{{ - \dfrac{1}{2}\left( {2\vartheta + 1} \right) + \dfrac{3}{2}}}{{\left( {{\vartheta ^2} + \vartheta + 1} \right)}}} d\vartheta = \int {\dfrac{{dx}}{x}} \\
   \Rightarrow - \dfrac{1}{2}\int {\dfrac{{\left( {2\vartheta + 1} \right)d\vartheta }}{{\left( {{\vartheta ^2} + \vartheta + 1} \right)}} + \dfrac{3}{2}\int {\dfrac{{d\vartheta }}{{\left( {{\vartheta ^2} + \vartheta + 1} \right)}}} } = \int {\dfrac{{dx}}{x}} \\
 \]
\[
   \Rightarrow - \dfrac{1}{2}\ell n\left| {{\vartheta ^2} + \vartheta + 1} \right| + \dfrac{3}{2}\int {\dfrac{{d\vartheta }}{{{{\left( {\vartheta + \dfrac{1}{2}} \right)}^2} + {{\left( {\dfrac{{\sqrt 3 }}{2}} \right)}^2}}} = \int {\dfrac{{dx}}{x}} } \\
   \Rightarrow - \dfrac{1}{2}\ell n\left| {{\vartheta ^2} + \vartheta + 1} \right| + \dfrac{3}{2} \times \dfrac{2}{{\sqrt 3 }}{\tan ^{ - 1}}\left( {\dfrac{{\vartheta + \dfrac{1}{2}}}{{\dfrac{{\sqrt 3 }}{2}}}} \right) = \ell n|x| + C \\
 \]
\[
   \Rightarrow - \dfrac{1}{2}\ell n\left| {{\vartheta ^2} + \vartheta + 1} \right| + \sqrt 3 {\tan ^{ - 1}}\left( {\dfrac{{2\vartheta + 1}}{{\sqrt 3 }}} \right) = \ell n\left| x \right| + C \\
    \because y = \vartheta x \\
   \Rightarrow \vartheta = \dfrac{y}{x} \\
   \Rightarrow - \dfrac{1}{2}\ell n\left| {\dfrac{{{y^2}}}{{{x^2}}} + \dfrac{y}{x} + 1} \right| + \sqrt 3 {\tan ^{ - 1}}\left( {\dfrac{{2y + x}}{{\sqrt 3 x}}} \right) = \ell n\left| x \right| + C \\
 \]
Now, we need to find the value of C
\[\because y = 0\]when \[x = 1\] , we have
\[
   - \dfrac{1}{2}\ell n\left| {0 + 0 + 1} \right| + \sqrt 3 {\tan ^{ - 1}}\left( {\dfrac{1}{{\sqrt 3 }}} \right) = \ell n\left| 1 \right| + C \\
   \Rightarrow - \dfrac{1}{2}\ell n\left| 1 \right| + \sqrt 3 {\tan ^{ - 1}}\left( {\dfrac{1}{{\sqrt 3 }}} \right) = \ell n\left| 1 \right| + C \\
   \Rightarrow 0 + \sqrt 3 \left( {\dfrac{\pi }{6}} \right) = 0 + C \\
   \Rightarrow C = \dfrac{{\sqrt 3 \pi }}{6} \\
  \,\,\,\,\,\therefore V = \dfrac{\pi }{{\sqrt 2 }} \\
 \]
Therefore, the particular solution of the given differential equation becomes
\[
   = \dfrac{1}{2}\ell n\left| {\dfrac{{{y^2}}}{{{x^2}}} + \dfrac{y}{x} + 1} \right| + \sqrt 3 {\tan ^{ - 1}}\left( {\dfrac{{2y + x}}{{\sqrt 3 x}}} \right) = \ell n\left| x \right| + \dfrac{\pi }{{\sqrt 2 }} \\
   \Rightarrow - \dfrac{1}{2}\ell n\left| {\dfrac{{{y^2} + xy + {x^2}}}{{{x^2}}}} \right| + \sqrt 3 {\tan ^{ - 1}}\left( {\dfrac{{2y + x}}{{\sqrt 3 x}}} \right) = \ell n\left| x \right| + \dfrac{\pi }{{\sqrt 2 }} \\
   \Rightarrow - \dfrac{1}{2}\ell n\left| {{x^2} + {y^2} + xy} \right| + \dfrac{1}{2}\ell n\left| {{x^2}} \right| + \sqrt 3 {\tan ^{ - 1}}\left( {\dfrac{{2y + x}}{{\sqrt 3 x}}} \right) = \ell n\left| x \right| + \dfrac{\pi }{{\sqrt 2 }} \\
   \Rightarrow - \dfrac{1}{2}\ell n\left| {{x^2} + {y^2} + xy} \right| + \dfrac{1}{2}\ell n\left( {{x^2}} \right) + \sqrt 3 {\tan ^{ - 1}}\left( {\dfrac{{2y + x}}{{\sqrt 3 x}}} \right) = \ell n\left| x \right| + \dfrac{\pi }{{\sqrt 2 }} \\
   \Rightarrow - \dfrac{1}{2}\ell n\left| {{x^2} + {y^2} + xy} \right|\ell n\left| x \right| + \sqrt 3 {\tan ^{ - 1}}\left( {\dfrac{{2y + x}}{{\sqrt 3 x}}} \right) = \ell n\left| x \right| + \dfrac{\pi }{{\sqrt 2 }} \\
   \Rightarrow - \dfrac{1}{2}\ell n\left| {{x^2} + {y^2} + xy} \right| + \sqrt 3 {\tan ^{ - 1}}\left( {\dfrac{{2y + x}}{{\sqrt 3 x}}} \right) = \dfrac{\pi }{{\sqrt 2 }} \\
 \]
Hence, the final solution is
\[ \Rightarrow - \dfrac{1}{2}\ell n\left| {{x^2} + {y^2} + xy} \right| + \sqrt 3 {\tan ^{ - 1}}\left( {\dfrac{{2y + x}}{{\sqrt 3 x}}} \right) = \dfrac{\pi }{{\sqrt 2 }}\]

Note: The students must keep in mind all the formulas of integration along with some standard forms of the differential equation. The problems of integration often require the similar kind of algorithm to be solved. Students are suggested to practice these kinds of questions as much as it is possible.