Mathematics

Solve the following differential equation.
$$\left( \cfrac { x+y-1 }{ x+y-2 }  \right) \dfrac { dy }{ dx } =\cfrac { x+y+1 }{ x+y+2 } $$


SOLUTION

Consider the given differential equation.

$$\left( \dfrac{x+y-1}{x+y-2} \right)\dfrac{dy}{dx}=\dfrac{x+y+1}{x+y+2}$$

 

Let,

$$ u=x+y $$

$$ \dfrac{du}{dx}=1+\dfrac{dy}{dx} $$

 

Therefore,

$$ \left( \dfrac{x+y-1}{x+y-2} \right)\dfrac{dy}{dx}=\dfrac{x+y+1}{x+y+2} $$

$$ \dfrac{u-1}{u-2}\left( \dfrac{du}{dx}-1 \right)=\dfrac{u+1}{u+2} $$

$$ \left( \dfrac{u-1}{u-2} \right)\dfrac{du}{dx}=\dfrac{u+1}{u+2}+\dfrac{u-1}{u-2} $$

$$ \left( \dfrac{u-1}{u-2} \right)\dfrac{du}{dx}=\dfrac{{{u}^{2}}+u-2u-2+{{u}^{2}}-u+2u-2}{\left( u-2 \right)\left( u+2 \right)} $$

$$ \left( \dfrac{u-1}{u-2} \right)\dfrac{du}{dx}=\dfrac{2\left( {{u}^{2}}-2 \right)}{\left( u-2 \right)\left( u+2 \right)} $$

$$ \dfrac{{{u}^{2}}-4}{{{u}^{2}}-2}du=2dx $$

 

Integrate both the sides.

$$ \int{\dfrac{{{u}^{2}}-4}{{{u}^{2}}-2}du}=\int{2dx} $$

$$ u+\dfrac{\ln \left( \dfrac{\left| u+\sqrt{2} \right|}{\left| u-\sqrt{2} \right|} \right)}{\sqrt{2}}=2x+c $$

$$ x+y+\dfrac{1}{\sqrt{2}}\ln \left( \dfrac{\left| x+y+\sqrt{2} \right|}{\left| x+y-\sqrt{2} \right|} \right)=2x+c $$

$$ x+y+\dfrac{1}{\sqrt{2}}\ln \left( \dfrac{\left| x+y+\sqrt{2} \right|}{\left| x+y-\sqrt{2} \right|} \right)+C=0 $$

 

Hence, this is the required solution of the integral.
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