For each of the differential equation in find the particular solu…

Question

For each of the differential equation in find the particular solution satisfying the given condition:$(\text{x}+\text{y})\ \text{dy}+(\text{x}-\text{y})\ \text{dx}=0;\ \text{y}=1\ \text{when}\ \text{x}=1$

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Answer

Given: Differential equation $(\text{x}+\text{y})\ \text{dy}+(\text{x}-\text{y})\ \text{dx}=0;\ \text{y}=1\ \text{when}\ \text{x}=1\ \ ...(\text{i})$ $\Rightarrow\ \ (\text{x}+\text{y})\ \text{dy}+(\text{x}-\text{y})\ \text{dx}=0\ \ $$\Rightarrow\ \ \frac{\text{dy}}{\text{dx}}=\frac{-(\text{x}-\text{y})}{(\text{x}+\text{y})}\ \ \Rightarrow\ \ \frac{\text{dy}}{\text{dx}}=\frac{\text{y}-\text{x}}{\text{y}+\text{x}}$ $\Rightarrow\ \ \frac{\text{dy}}{\text{dx}}=\frac{\text{x}\Big(\frac{\text{y}}{\text{x}}-1\Big)}{\text{x}\Big(\frac{\text{y}}{\text{x}}+1\Big)}$ $\ \ \Rightarrow\ \ \frac{\text{dy}}{\text{dx}}=\frac{\Big(\frac{\text{y}}{\text{x}}-1\Big)}{\Big(\frac{\text{y}}{\text{x}}+1\Big)}=f\Big(\frac{\text{y}}{\text{x}}\Big)\ \ .....(\text{ii})$ Therefore the given differential equation is homogeneous because each coefficient of dx and dy is same i.e., degree 2.$\text{Putting}\frac{\text{y}}{\text{x}}=\text{v}\ \ \Rightarrow\ \ \text{y}=\text{vx}\ \ $ $\Rightarrow\ \ \frac{\text{dy}}{\text{dx}}=\text{v}+\text{x}\frac{\text{dv}}{\text{dx}}$
$\text{Putting these values of}\ \frac{\text{y}}{\text{x}}\ \text{and}\ \frac{\text{dy}}{\text{dx}}\ \text{in eq. (ii), we have}$ $\text{v}+\text{x}\frac{\text{dv}}{\text{dx}}=\frac{\text{v}-1} {\text{v}+1}\ \ \Rightarrow\ \ \text{x}\frac{\text{dv}}{\text{dx}}=\frac{\text{v}-1} {\text{v}+1}-\text{v}\ \ $ $\Rightarrow\ \ \text{x}\frac{\text{dv}}{\text{dx}}=\frac{\text{v}-1-\text{v}^2-\text{v}} {\text{v}+1}=\frac{-\text{v}^2-1}{\text{v}+1}$ $\Rightarrow\ \ \text{x}\frac{\text{dv}}{\text{dx}}=\frac{\text{v}^2+1} {\text{v}+1}\ \ \Rightarrow\ \ \text{x}(\text{v}+1)\ \text{dv}=-(\text{v}^2+1)\ \text{dx}$ $\Rightarrow\ \ \frac{\text{v}+1} {\text{v}^2+1}\text{dv}=-\frac{\text{dx}}{\text{x}}\ \ \big[\text{Separating variables}\big]$ $\text{Integrating both sides,}$$\ \ \int\frac{\text{v}} {\text{v}^2+1}\cdot\text{dv}+ \int\frac{1} {\text{v}^2+1}\text{dv}=-\int\frac{1}{\text{x}}\text{dx}$ $\Rightarrow\ \ \frac{1}{2}\int\frac{2\text{v}}{\text{v}^2+1}\ \text{dv}+\tan^{-1}\ \text{v}=-\log\text{x}+\text{c}\ \ $ $\Rightarrow\ \ \frac{1}{2}\log(\text{v}^2+1)+\tan^{-1}\text{v}=-\log\text{x}+\text{c}$ $\text{Now putting v}=\frac{\text{y}}{\text{x}},\ \ $ $\frac{1}{2}\log\bigg(\Big(\frac{\text{y}}{\text{x}}\Big)^2+1\bigg)+\tan^{-1}\frac{\text{y}}{\text{x}}=-\log\text{x}+\text{c}$ $\Rightarrow\ \ \frac{1}{2}\log\bigg(\frac{\text{y}^2+\text{x}^2}{\text{x}^2}\bigg)+\tan^{-1}\frac{\text{y}}{\text{x}}=-\log\ \text{x}+\text{c}$ $\Rightarrow\ \ \frac{1}{2}\log(\text{y}^2+\text{x}^2)-\frac{1}{2}\log\ \text{x}^2+\tan^{-1}\frac{\text{y}}{\text{x}}=-\log\ \text{x}+\text{c}$ $\Rightarrow\ \ \frac{1}{2}\log\big(\text{y}^2+\text{x}^2\big)-\frac{1}{2}.2\log\ \text{x}+\tan^{-1}\frac{\text{y}}{\text{x}}=-\log\ \text{x}+\text{c}$ $\Rightarrow\ \ \frac{1}{2}\log\big(\text{y}^2+\text{x}^2\big)+\tan^{-1}\frac{\text{y}}{\text{x}}=\text{c}\ \ ....\text{(iii)}$Now again given y = 1 when x = 1, therefore putting these values in eq. (iii),
$\frac{1}{2}\log\big(1+1\big)+\tan^{-1}1=\text{c}\ \ \Rightarrow\ \ \text{c}=\frac{1}{2}\log2+\frac{\pi}{4}$ Putting this value of c in eq. (iii), we get $\frac{1}{2}\log\big(\text{y}^2+\text{x}^2\big)+\tan^{-1}\frac{\text{y}}{\text{x}}=\frac{1}{2}\log2+\frac{\pi}{4}$ $\Rightarrow\ \ \log\big(\text{x}^2+\text{y}^2\big)+2\tan^{-1}\frac{\text{y}}{\text{x}}=\log2+\frac{\pi}{4}$