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Question: Solve the given differential equation \(\dfrac{{dy}}{{dx}} = \dfrac{{{x^2} + xy}}{{{x^2} + {y^2}}}\)...

Solve the given differential equation dydx=x2+xyx2+y2\dfrac{{dy}}{{dx}} = \dfrac{{{x^2} + xy}}{{{x^2} + {y^2}}}
(a) c(x−y)2/3(x2+xy+y2)1/6=exp⁡[−13tan⁡−1x+2yx3] where exp x = ex\left( a \right){\text{ }}\dfrac{{c{{\left( {x - y} \right)}^{2/3}}}}{{{{\left( {{x^2} + xy + {y^2}} \right)}^{1/6}}}} = \exp \left[ { - \dfrac{1}{{\sqrt 3 }}{{\tan }^{ - 1}}\dfrac{{x + 2y}}{{x\sqrt 3 }}} \right]{\text{ where exp x = }}{{\text{e}}^x}
(b) c(x−y)4/3(x2+xy+y2)1/4=exp⁡[−13sec⁡−1x−2yx3] where exp x = ex\left( b \right){\text{ c}}{\left( {x - y} \right)^{4/3}}{\left( {{x^2} + xy + {y^2}} \right)^{1/4}} = \exp \left[ { - \dfrac{1}{{\sqrt 3 }}{{\sec }^{ - 1}}\dfrac{{x - 2y}}{{x\sqrt 3 }}} \right]{\text{ where exp x = }}{{\text{e}}^x}
(c) c(x−y)2/3(x2+xy+y2)1/6=exp⁡[−13sec⁡−1x−2yx5] where exp x = ex\left( c \right){\text{ }}\dfrac{{c{{\left( {x - y} \right)}^{2/3}}}}{{{{\left( {{x^2} + xy + {y^2}} \right)}^{1/6}}}} = \exp \left[ { - \dfrac{1}{{\sqrt 3 }}{{\sec }^{ - 1}}\dfrac{{x - 2y}}{{x\sqrt 5 }}} \right]{\text{ where exp x = }}{{\text{e}}^x}
(d) None of these\left( d \right){\text{ None of these}}

Explanation

Solution

Since we have to solve this and by looking at the question we can say that it is a homogeneous differential equation. So we will put y=vxy = vxand then after equating we will integrate it and by using the partial fraction we will get the values and on further integration, we will get the solution.

Complete step-by-step answer:
Given, we have the function
dydx=x2+xyx2+y2\dfrac{{dy}}{{dx}} = \dfrac{{{x^2} + xy}}{{{x^2} + {y^2}}}
Since this is a homogeneous differential equation.
So for solving this first of all we will put y=vxy = vx
On differentiating the above term we get
⇒dydx=v+xdvdx\Rightarrow \dfrac{{dy}}{{dx}} = v + x\dfrac{{dv}}{{dx}}
So the above equation will become
⇒v+xdvdx=1+v1+v2\Rightarrow v + x\dfrac{{dv}}{{dx}} = \dfrac{{1 + v}}{{1 + {v^2}}}
And on solving the equation, we get
⇒xdvdx=1−v31+v2\Rightarrow x\dfrac{{dv}}{{dx}} = \dfrac{{1 - {v^3}}}{{1 + {v^2}}}
And it can also be written as
⇒1+v21−v3dv=dxx\Rightarrow \dfrac{{1 + {v^2}}}{{1 - {v^3}}}dv = \dfrac{{dx}}{x}
Now, on integrating both the sides we get
⇒∫1+v2(1−v)(1+v2+v)dv=∫dxx\Rightarrow \int {\dfrac{{1 + {v^2}}}{{\left( {1 - v} \right)\left( {1 + {v^2} + v} \right)}}} dv = \int {\dfrac{{dx}}{x}}, let’s name it equation 11
Now on resolving ⇒1+v2(1−v)(1+v2+v) \Rightarrow \dfrac{{1 + {v^2}}}{{\left( {1 - v} \right)\left( {1 + {v^2} + v} \right)}} into the partial fractions, we get
⇒1+v2(1−v)(1+v2+v)=A(1−v)+Bv+C(1+v2+v)\Rightarrow \dfrac{{1 + {v^2}}}{{\left( {1 - v} \right)\left( {1 + {v^2} + v} \right)}} = \dfrac{A}{{\left( {1 - v} \right)}} + \dfrac{{Bv + C}}{{\left( {1 + {v^2} + v} \right)}}, and we will name its equation22.
So on solving the partial fractions, we get
⇒1+v2=A(1+v2+v)+(Bv+c)(1−v)\Rightarrow 1 + {v^2} = A\left( {1 + {v^2} + v} \right) + \left( {Bv + c} \right)\left( {1 - v} \right)
And on comparing the above equation, we get
⇒A−B=1\Rightarrow A - B = 1 ; A+C=1A + C = 1; A+B−C=0A + B - C = 0
So now on solving these equations, we get
⇒A=23 ; B = −13 ; C = 13\Rightarrow A = \dfrac{2}{3}{\text{ ; B = }}\dfrac{{ - 1}}{3}{\text{ ; C = }}\dfrac{1}{3}
So now we will substitute these values in the equation22, we get
⇒1+v2(1−v)(1+v2+v)=23(1−v)−13(v−1)(1+v2+v)\Rightarrow \dfrac{{1 + {v^2}}}{{\left( {1 - v} \right)\left( {1 + {v^2} + v} \right)}} = \dfrac{2}{{3\left( {1 - v} \right)}} - \dfrac{1}{3}\dfrac{{\left( {v - 1} \right)}}{{\left( {1 + {v^2} + v} \right)}}
Therefore the equation 11 will become
⇒∫23(1−v)dv−∫13(v−1)(1+v2+v)dv=∫dxx\Rightarrow \int {\dfrac{2}{{3\left( {1 - v} \right)}}dv - \int {\dfrac{1}{3}\dfrac{{\left( {v - 1} \right)}}{{\left( {1 + {v^2} + v} \right)}}} } dv = \int {\dfrac{{dx}}{x}}
Now on applying the integral formula, we get
⇒23log⁡∣1−v∣−16∫(2v+1−3)(1+v2+v)dv=log⁡x+log⁡c\Rightarrow \dfrac{2}{3}\log \left| {1 - v} \right| - \dfrac{1}{6}\int {\dfrac{{\left( {2v + 1 - 3} \right)}}{{\left( {1 + {v^2} + v} \right)}}} dv = \log x + \log c
Now, on further integrating more, we get
⇒23log⁡∣1−v∣−16∫(2v+1)(1+v2+v)dv+12∫1(v+12)2+(32)2dv=log⁡x+log⁡c\Rightarrow \dfrac{2}{3}\log \left| {1 - v} \right| - \dfrac{1}{6}\int {\dfrac{{\left( {2v + 1} \right)}}{{\left( {1 + {v^2} + v} \right)}}} dv + \dfrac{1}{2}\int {\dfrac{1}{{{{\left( {v + \dfrac{1}{2}} \right)}^2} + {{\left( {\dfrac{{\sqrt 3 }}{2}} \right)}^2}}}dv} = \log x + \log c
Now on solving this, we get
⇒23log⁡∣1−yx∣−16log⁡∣v2+v+1∣+13tan⁡−1(2v+13)=log⁡x+log⁡c\Rightarrow \dfrac{2}{3}\log \left| {1 - \dfrac{y}{x}} \right| - \dfrac{1}{6}\log \left| {{v^2} + v + 1} \right| + \dfrac{1}{{\sqrt 3 }}{\tan ^{ - 1}}\left( {\dfrac{{2v + 1}}{{\sqrt 3 }}} \right) = \log x + \log c
Combining and solving the same term together, we get
⇒23log⁡∣x−y∣−16log⁡∣y2+xy+x2∣+23log⁡x+13tan⁡−1(2y+xx3)=53log⁡x+log⁡c\Rightarrow \dfrac{2}{3}\log \left| {x - y} \right| - \dfrac{1}{6}\log \left| {{y^2} + xy + {x^2}} \right| + \dfrac{2}{3}\log x + \dfrac{1}{{\sqrt 3 }}{\tan ^{ - 1}}\left( {\dfrac{{2y + x}}{{x\sqrt 3 }}} \right) = \dfrac{5}{3}\log x + \log c
Now solving for the log by using the log property, we get
⇒log⁡c(x−y)2/3x(x2+xy+y2)1/6=−13tan⁡−12y+xx3\Rightarrow \log \dfrac{{c{{\left( {x - y} \right)}^{2/3}}}}{{x{{\left( {{x^2} + xy + {y^2}} \right)}^{1/6}}}} = - \dfrac{1}{{\sqrt 3 }}{\tan ^{ - 1}}\dfrac{{2y + x}}{{x\sqrt 3 }}
On removing the log, we get
⇒c(x−y)2/3x(x2+xy+y2)1/6=exp⁡[−13tan⁡−12y+xx3]\Rightarrow \dfrac{{c{{\left( {x - y} \right)}^{2/3}}}}{{x{{\left( {{x^2} + xy + {y^2}} \right)}^{1/6}}}} = \exp \left[ { - \dfrac{1}{{\sqrt 3 }}{{\tan }^{ - 1}}\dfrac{{2y + x}}{{x\sqrt 3 }}} \right]
Therefore, the option (a)\left( a \right) will be correct.

Note: So for solving such type of problem we should have to remember the formulas which are going to be used and with practice, we can develop the skills of using the methods for solving this type of problem easily.