Find the particular solution of the differential equation (x-y) dy dx =x +2y, given that when x = 1, y = 0.

Consider the given equation (x-y) dy dx =x +2y This is a homogeneous equation. Substituting y = vx and dy dx = (v + x dv dx )In the above equation, we have, (x-vx) (v + x dv dx )=x +2vx (1-v) (v + x dv dx )=1+2v v + x dv dx = 1+2v 1-v x dv dx = 1+2v 1-v -v x dv dx = 1+2v-v(1-v) 1-v x dv dx = 1+2v-v+v^2 1-v x dv dx = 1+v+v^2 1-v (1-v)dv (1+v+v^2) = dx x Integrating both sides, we have, (1-v)dv (1+v+v^2) = dx x 32 dv (1+v+v^2) - 12 (2v+1)dv (1+v+v^2) = dx x 32 dv v^2+14+v+34 - 12 (2v+1)dv (1+v+v^2) = dx x 32 dv (v+12 )^2+ ( 32 )^2 - 12 (2v+1)dv (1+v+v^2) = dx x 32 1 3 2 ^ -1 v+1 2 3 2 -12 (1+v+v^2)= +C 3 ^ -1 2v+1 3 -12 (1+v+v^2)= +C 3 ^ -1 2 ( y x )+1 3 -1 2 (1+ ( y x )+ ( y x )^2 )= +C (1) Given that when x = 1, y = 0 Substituting the values, in the above equation, we get, 3 ^ -1 2 0+1 3 -12 (1+0+0^2)= 1+C 3 ^ -1 1 3 -1 2 0=0+C C= 3 6 C= 2 3 Thus equation (1) becomes, 3 ^ -1 2 ( y x )+1 3 -12 (1+ ( y x )+ ( y x )^2 )= + 2 3 3 ^ -1 2y + x x 3 - 2 3 = +12 (1+ ( y x )+ ( y x )^2 ) 2 3 ^ -1 2y + x x 3 - 3 = ^2+ ( x^2+xy+y^2 x^2 ) 2 3 ^ -1 2y + x x 3 - 3 = (x^2+xy+y^2)

Find the particular solution of the differential equation (x-y) d…

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Question

Find the particular solution of the differential equation $(\text{x}-\text{y})\frac{\text{dy}}{\text{dx}}=\text{x +2y},$ given that when x = 1, y = 0.

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Answer

Consider the given equation $(\text{x}-\text{y})\frac{\text{dy}}{\text{dx}}=\text{x +2y}$ This is a homogeneous equation. Substituting y = vx and $\frac{\text{dy}}{\text{dx}}=\Big(\text{v + x}\frac{\text{dv}}{\text{dx}}\Big)$In the above equation, we have,
$(\text{x}-\text{vx})\Big(\text{v + x}\frac{\text{dv}}{\text{dx}}\Big)=\text{x +2vx}$ $\Rightarrow\ (1-\text{v})\Big(\text{v + x}\frac{\text{dv}}{\text{dx}}\Big)=1+2\text{v}$ $\Rightarrow\ \text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{1+2\text{v}}{1-\text{v}}$ $\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=\frac{1+2\text{v}}{1-\text{v}}-\text{v}$ $\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=\frac{1+2\text{v}-\text{v}(1-\text{v})}{1-\text{v}}$ $\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=\frac{1+2\text{v}-\text{v}+\text{v}^2}{1-\text{v}}$ $\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=\frac{1+\text{v}+\text{v}^2}{1-\text{v}}$ $\Rightarrow\ \frac{(1-\text{v})\text{dv}}{(1+\text{v}+\text{v}^2)}=\frac{\text{dx}}{\text{x}}$ Integrating both sides, we have, $\Rightarrow\ \int\frac{(1-\text{v})\text{dv}}{(1+\text{v}+\text{v}^2)}=\int\frac{\text{dx}}{\text{x}}$ $\Rightarrow\ \frac{3}2\int\frac{\text{dv}}{(1+\text{v}+\text{v}^2)}-\int\frac{1}2\frac{(2\text{v}+1)\text{dv}}{(1+\text{v}+\text{v}^2)}=\int\frac{\text{dx}}{\text{x}}$ $\Rightarrow\ \frac{3}2\int\frac{\text{dv}}{\text{v}^2+\frac{1}4+\text{v}+\frac{3}4}-\int\frac{1}2\frac{(2\text{v}+1)\text{dv}}{(1+\text{v}+\text{v}^2)}=\int\frac{\text{dx}}{\text{x}}$ $\Rightarrow\ \frac{3}2\int\frac{\text{dv}}{\big(\text{v}+\frac{1}2\big)^2+\big(\frac{\sqrt3}2\big)^2}-\int\frac{1}2\frac{(2\text{v}+1)\text{dv}}{(1+\text{v}+\text{v}^2)}=\int\frac{\text{dx}}{\text{x}}$ $\Rightarrow\ \frac{3}2\times\frac{1}{\frac{\sqrt3}{2}}\tan^{-1}\frac{\text{v}+\frac{1}{2}}{\frac{\sqrt3}{2}}-\frac{1}2\log(1+\text{v}+\text{v}^2)=\log\text{x}+\text{C}$ $\Rightarrow\ \sqrt3\tan^{-1}\frac{2\text{v}+1}{\sqrt3}-\frac{1}2\log(1+\text{v}+\text{v}^2)=\log\text{x}+\text{C}$ $\Rightarrow\ \sqrt3\tan^{-1}\frac{2\big(\frac{\text{y}}{\text{x}}\big)+1}{\sqrt3}-\frac{1}{2}\log\Big(1+\Big(\frac{\text{y}}{\text{x}}\Big)+\Big(\frac{\text{y}}{\text{x}}\Big)^2\Big)=\log\text{x}+\text{C}\ \dots(1)$ Given that when x = 1, y = 0 Substituting the values, in the above equation, we get, $\Rightarrow\ \sqrt3\tan^{-1}\frac{2\times0+1}{\sqrt3}-\frac{1}2\log(1+0+0^2)=\log1+\text{C}$ $\Rightarrow\ \sqrt3\tan^{-1}\frac{1}{\sqrt3}-\frac{1}{2}\times0=0+\text{C}$ $\Rightarrow\ \text{C}=\sqrt3\times\frac{\pi}{6}$ $\Rightarrow\ \text{C}=\frac{\pi}{2\sqrt3}$ Thus equation (1) becomes, $\sqrt3\tan^{-1}\frac{2\big(\frac{\text{y}}{\text{x}}\big)+1}{\sqrt3}-\frac{1}2\log\Big(1+\Big(\frac{\text{y}}{\text{x}}\Big)+\Big(\frac{\text{y}}{\text{x}}\Big)^2\Big)=\log\text{x}+\frac{\pi}{2\sqrt3}$ $\Rightarrow\ \sqrt3\tan^{-1}\frac{2\text{y + x}}{\text{x}\sqrt3}-\frac{\pi}{2\sqrt3}=\log\text{x}+\frac{1}2\log\Big(1+\Big(\frac{\text{y}}{\text{x}}\Big)+\Big(\frac{\text{y}}{\text{x}}\Big)^2\Big)$ $\Rightarrow\ 2\sqrt3\tan^{-1}\frac{2\text{y + x}}{\text{x}\sqrt3}-\frac{\pi}{\sqrt3}=\log\text{x}^2+\log\Big(\frac{\text{x}^2+\text{xy}+\text{y}^2}{\text{x}^2}\Big)$ $\Rightarrow\ 2\sqrt3\tan^{-1}\frac{2\text{y + x}}{\text{x}\sqrt3}-\frac{\pi}{\sqrt3}=\log(\text{x}^2+\text{xy}+\text{y}^2)$