Question

What is the solution of the differential equation \[{\left( {x + y} \right)^2}\dfrac{{dy}}{{dx}} = {a^2}\]?
A. \[{\left( {x + y} \right)^2} = \dfrac{{{a^2}}}{2}x + c\]
B. \[{\left( {x + y} \right)^2} = {a^2}x + c\]
C. \[{\left( {x + y} \right)^2} = 2{a^2}x + c\]
D. None of these

Answer 1

Hint: The given differential equation is a homogeneous equation. Thus, to solve the given differential equation, we will use the substitution method.

Formula Used: Integration formula:
\[\int {\dfrac{1}{{{a^2} + {x^2}}}dx} = \dfrac{1}{a}{\tan ^{ - 1}}\dfrac{x}{a} + c\]
Homogeneous equation: The function \[f\left( {x,y} \right)\] is homogeneous with degree n if \[f\left( {\lambda x,\lambda y} \right) = {\lambda ^n}f\left( {x,y} \right)\] where \[\lambda \] is nonzero constant.

Complete step by step solution: Given differential equation is
\[{\left( {x + y} \right)^2}\dfrac{{dy}}{{dx}} = {a^2}\]
Rewrite the above equation:
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{{a^2}}}{{{{\left( {x + y} \right)}^2}}}\]
Compare equation (i) with \[\dfrac{{dy}}{{dx}} = f\left( {x,y} \right)\]
\[f\left( {x,y} \right) = \dfrac{{{a^2}}}{{{{\left( {x + y} \right)}^2}}}\]
Putting \[x = \lambda x\] and \[y = \lambda y\]
\[f\left( {\lambda x,\lambda y} \right) = \dfrac{{{a^2}}}{{{{\left( {\lambda x + \lambda y} \right)}^2}}}\]
\[ \Rightarrow f\left( {\lambda x,\lambda y} \right) = \dfrac{{{a^2}}}{{{\lambda ^2}{{\left( {x + y} \right)}^2}}}\]
\[ \Rightarrow f\left( {\lambda x,\lambda y} \right) = {\lambda ^{ - 2}}f\left( {x,y} \right)\]
The given differential equation is a homogenous with degree 2.
Since given differential equation is a homogenous equation, we will apply substitution method.
Assume that,
\[x + y = z\]
Differentiate both sides with respect to x:
\[ \Rightarrow 1 + \dfrac{{dy}}{{dx}} = \dfrac{{dz}}{{dx}}\]
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{dz}}{{dx}} - 1\]
Substitute \[x + y = z\] and \[\dfrac{{dy}}{{dx}} = \dfrac{{dz}}{{dx}} - 1\] in the given differential equation:
\[{z^2}\left( {\dfrac{{dz}}{{dx}} - 1} \right) = {a^2}\]
\[ \Rightarrow \left( {\dfrac{{dz}}{{dx}} - 1} \right) = \dfrac{{{a^2}}}{{{z^2}}}\]
\[ \Rightarrow \dfrac{{dz}}{{dx}} - 1 + 1 = \dfrac{{{a^2}}}{{{z^2}}} + 1\]
Add 1 on both sides
\[ \Rightarrow \dfrac{{dz}}{{dx}} = \dfrac{{{a^2} + {z^2}}}{{{z^2}}}\]
\[ \Rightarrow \dfrac{{{z^2}dz}}{{{a^2} + {z^2}}} = dx\]
Add and subtract \[{a^2}\] with numerator:
\[ \Rightarrow \dfrac{{{z^2} - {a^2} + {a^2}}}{{{a^2} + {z^2}}}dz = dx\]
\[ \Rightarrow \left( {1 - \dfrac{{{a^2}}}{{{a^2} + {z^2}}}} \right)dz = dx\]
Taking integration on both sides:
\[ \Rightarrow \int {dz} - \int {\dfrac{{{a^2}}}{{{a^2} + {z^2}}}dz} = \int {dx} \]
\[ \Rightarrow z - {a^2} \cdot \dfrac{1}{a}{\tan ^{ - 1}}\dfrac{z}{a} = x + c\]
\[ \Rightarrow z - a{\tan ^{ - 1}}\dfrac{z}{a} = x + c\]
Substitute the value of z:
\[ \Rightarrow x + y - a{\tan ^{ - 1}}\left( {\dfrac{{x + y}}{a}} \right) = x + c\]
\[ \Rightarrow x + y - x = a{\tan ^{ - 1}}\left( {\dfrac{{x + y}}{a}} \right) + c\]
\[ \Rightarrow y = a{\tan ^{ - 1}}\left( {\dfrac{{x + y}}{a}} \right) + c\]

Option ‘D’ is correct

Note: Students often do a mistake to integrate \[\int {\dfrac{1}{{{a^2} + {x^2}}}dx} \]. They applied an incorrect formula that is \[\int {\dfrac{1}{{{a^2} + {x^2}}}dx} = {\tan ^{ - 1}}\dfrac{x}{a} + c\]. They forgot to multiply \[\dfrac{1}{a}\]. The correct formula is \[\int {\dfrac{1}{{{a^2} + {x^2}}}dx} = \dfrac{1}{a}{\tan ^{ - 1}}\dfrac{x}{a} + c\].