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Question-123009




Question Number 123009 by aurpeyz last updated on 21/Nov/20
Commented by aurpeyz last updated on 21/Nov/20
pls help with question 1
$${pls}\:{help}\:{with}\:{question}\:\mathrm{1} \\ $$
Answered by Dwaipayan Shikari last updated on 21/Nov/20
(x+y)^2 (dy/dx)=a^2            x+y=v⇒(dy/dx)=(dv/dx)−1  v^2 ((dv/dx)−1)=a^2   v^2 (dv/dx)=v^2 +a^2 ⇒∫1−(a^2 /(v^2 +a^2 ))dv=∫dx  ⇒v−atan^(−1) (v/a)=x  ⇒x+y−atan^(−1) ((x+y)/a)=x  ⇒(y/a)=tan^(−1) ((x+y)/a)⇒atan(y/a)−y=x
$$\left({x}+{y}\right)^{\mathrm{2}} \frac{{dy}}{{dx}}={a}^{\mathrm{2}} \:\:\:\:\:\:\:\:\:\:\:{x}+{y}={v}\Rightarrow\frac{{dy}}{{dx}}=\frac{{dv}}{{dx}}−\mathrm{1} \\ $$$${v}^{\mathrm{2}} \left(\frac{{dv}}{{dx}}−\mathrm{1}\right)={a}^{\mathrm{2}} \\ $$$${v}^{\mathrm{2}} \frac{{dv}}{{dx}}={v}^{\mathrm{2}} +{a}^{\mathrm{2}} \Rightarrow\int\mathrm{1}−\frac{{a}^{\mathrm{2}} }{{v}^{\mathrm{2}} +{a}^{\mathrm{2}} }{dv}=\int{dx} \\ $$$$\Rightarrow{v}−{atan}^{−\mathrm{1}} \frac{{v}}{{a}}={x} \\ $$$$\Rightarrow{x}+{y}−{atan}^{−\mathrm{1}} \frac{{x}+{y}}{{a}}={x} \\ $$$$\Rightarrow\frac{{y}}{{a}}={tan}^{−\mathrm{1}} \frac{{x}+{y}}{{a}}\Rightarrow{atan}\frac{{y}}{{a}}−{y}={x} \\ $$
Commented by aurpeyz last updated on 22/Nov/20
thats like homogenous equation?
$${thats}\:{like}\:{homogenous}\:{equation}? \\ $$

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