Find a particular solution of the differential equation (x-y)(dx+dy)=dx-dy, given that y=-1, when x=0. (Hint: put x-y=t)

Given: Differential equation (x-y)(dx+dy)=dx-dy (x-y)dx+(x-y)dy=dx-dy (x-y)dx-dx+(x-y)dy+dy=0 (x-y-1)dx+(x-y+1)dy=0 (x-y-1)dx=-(x-y+1)dy dy dx =- (x-y-1) x-y+1 ...(i) Putting x-y=t 1- dy dx = dt dx dy dx = dt dx -1 dy dx = -dt dx +1 Putting this value in eq. (i), -dt dx +1=- ( t-1 t+1 ) -dt dx =-1- ( t-1 t+1 ) dt dx =1+ ( t-1 t+1 )= t+1+t-1 t+1 dt dx = 2t t+1 (t+1)dt=2t dx t+1 t dt=2 dx Integrating both sides, ( t+1 t )dt=2 1 dx ( t t +1 t )dt=2x+c (1+1 t )dt=2x+c t+ |t|=2x+c Putting x-y=t, x-y+ |x-y|=2x+c |x-y|=x+y+c .....(ii) Now putting y=-1, x=0 in eq. (ii), 1=0-1+c 0=-1+c c=1 Putting c=1 in eq. (ii), |x-y|=x+y+1

Find a particular solution of the differential equation (x-y)(dx+…

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Question

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

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Answer

Given: Differential equation $(\text{x}-\text{y})(\text{dx}+\text{dy})=\text{dx}-\text{dy}$
$\Rightarrow\ \ (\text{x}-\text{y})\text{dx}+(\text{x}-\text{y})\text{dy}=\text{dx}-\text{dy}$ $\Rightarrow\ \ (\text{x}-\text{y})\text{dx}-\text{dx}+(\text{x}-\text{y})\text{dy}+\text{dy}=0$
$\Rightarrow\ \ (\text{x}-\text{y}-1)\text{dx}+(\text{x}-\text{y}+1)\text{dy}=0$ $\Rightarrow\ \ (\text{x}-\text{y}-1)\text{dx}=-(\text{x}-\text{y}+1)\text{dy}$
$\Rightarrow\ \ \frac{\text{dy}}{\text{dx}}=-\frac{(\text{x}-\text{y}-1)}{\text{x}-\text{y}+1}\ \ ...\text{(i)}$
$\text{Putting x}-\text{y}=\text{t}\ \ \Rightarrow\ \ 1-\frac{\text{dy}}{\text{dx}}=\frac{\text{dt}}{\text{dx}}$ $\Rightarrow\ \ \frac{\text{dy}}{\text{dx}}=\frac{\text{dt}}{\text{dx}}-1\ \ \Rightarrow\ \ \frac{\text{dy}}{\text{dx}}=\frac{-\text{dt}}{\text{dx}}+1$
$\text{Putting this value in eq. (i),}\ \ \frac{-\text{dt}}{\text{dx}}+1=-\Big(\frac{\text{t}-1}{\text{t}+1}\Big)$ $ \Rightarrow\ \ \frac{-\text{dt}}{\text{dx}}=-1-\Big(\frac{\text{t}-1}{\text{t}+1}\Big)$
$\Rightarrow\ \ \frac{\text{dt}}{\text{dx}}=1+\Big(\frac{\text{t}-1}{\text{t}+1}\Big)=\frac{\text{t}+1+\text{t}-1}{\text{t}+1}$ $ \Rightarrow\ \ \frac{\text{dt}}{\text{dx}}=\frac{2\text{t}}{\text{t}+1}$
$\Rightarrow\ \ (\text{t}+1)\text{dt}=2\text{t}\ \text{dx}\ \ \Rightarrow\ \ \frac{\text{t}+1}{\text{t}}\text{dt}=2\ \text{dx}$
$\text{Integrating both sides,}\ \ \int\Big(\frac{\text{t}+1}{\text{t}}\Big)\text{dt}=2\int1\ \text{dx}$ $\Rightarrow\ \ \int\Big(\frac{\text{t}}{\text{t}}+\frac{1}{\text{t}}\Big)\text{dt}=2\text{x}+\text{c}$
$\Rightarrow\ \ \int\Big(1+\frac{1}{\text{t}}\Big)\text{dt}=2\text{x}+\text{c}$ $\Rightarrow\ \ \text{t}+\log|\text{t}|=2\text{x}+\text{c}$
$\text{Putting x}-\text{y}=\text{t},\ \ \text{x}-\text{y}+\log|\text{x}-\text{y}|=2\text{x}+\text{c}$ $\Rightarrow\ \ \log|\text{x}-\text{y}|=\text{x}+\text{y}+\text{c}\ \ .....\text{(ii)}$
$\text{Now putting y}=-1,\ \text{x}=0\ \text{in eq. (ii)},$ $\log1=0-1+\text{c}\ \ \Rightarrow\ \ 0=-1+\text{c}\ \ \Rightarrow\ \ \text{c}=1$
$\text{Putting c}=1\ \text{in eq. (ii)},\ \ \log|\text{x}-\text{y}|=\text{x}+\text{y}+1$

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