Question
For each of the differential equation in find the particular solution satisfying the given condition:
$\text{x}^2\ \text{dy}+(\text{xy}+\text{y}^2)\ \text{dx}=0;\ \text{y}=1\ \text{when}\ \text{x}=1$
✓
Answer
Given: Differential equation $\text{x}^2\ \text{dy}+(\text{xy}+\text{y}^2)\ \text{dx}=0$ $\Rightarrow\ \ \text{x}^2\ \text{dy}-(\text{xy}+\text{y}^2)\ \text{dx}\ \ $ $\Rightarrow\ \ \frac{\text{dy}}{\text{dx}}=-\frac{\text{y}(\text{x}+\text{y})}{\text{x}^2}=-\frac{\text{xy}\Big(1+\frac{\text{y}}{\text{x}}\Big)}{\text{x}^2}$ $\Rightarrow\ \ \frac{\text{dy}}{\text{dx}}=-\frac{\text{y}}{\text{x}}\Big(1+\frac{\text{y}}{\text{x}}\Big)=f\Big(\frac{\text{y}}{\text{x}}\Big)\ \ ...\text{(i)}$ Therefore the given differential equation is homogeneous. $\text{Putting}\frac{\text{y}}{\text{x}}=\text{v}\ \ \Rightarrow\ \ \text{y}=\text{vx}\ \ $ $\Rightarrow\ \ \frac{\text{dy}}{\text{dx}}=\text{v}+\text{x}\frac{\text{dv}}{\text{dx}}$ $\text{Putting these values of}\ \frac{\text{y}}{\text{x}}\ \text{and}\ \frac{\text{dv}}{\text{dx}}\ \text{in eq. (ii), we have}$ $\text{v}+\text{x}\frac{\text{dv}}{\text{dx}}=-\text{v}(1+\text{v})=-\text{v}-\text{v}^2\ \ $ $\Rightarrow\ \ \text{x}\frac{\text{dv}}{\text{dx}}=-\text{v}^2-2\text{v}$ $\Rightarrow\ \ \text{x}\frac{\text{dv}}{\text{dx}}=-\text{v}(\text{v}+2)\ \ \Rightarrow\ \ \frac{\text{dv}}{\text{v}(\text{v}+2)}=-\frac{\text{dx}}{\text{x}}$ $\text{Integrating both sides,}$ $\ \ \int\frac{1} {\text{v}(\text{v}+2)}\text{dv}=-\int\frac{1}{\text{x}}\text{dx}$ $\Rightarrow\ \ \frac{1}{2}\int\frac{2} {\text{v}(\text{v}+2)}\text{dv}=-\log|\text{x}|+\log|\text{c}|\ \ $ $\Rightarrow\ \ \frac{1}{2}\int\frac{(\text{v}+2)-\text{v}}{\text{v}(\text{v}+2)}\ \text{dv}=-\log|\text{x}|+\log|\text{c}|$ $\Rightarrow\ \ \int\Big(\frac{1}{\text{v}}-\frac{1}{\text{v+2}}\Big)\text{dv}=-2\log|\text{x}|+\log|\text{c}|$ $\Rightarrow\ \ \log\Big|\frac{\text{v}}{\text{v}+2}\Big|=\log\big|\text{x}^{-2}\big|+\log|\text{c}|$ $\Rightarrow\ \ \log\Big|\frac{\text{v}}{\text{v}+2}\Big|=\log\big|\text{cx}^{-2}\big|\ \ $ $\Rightarrow\ \ \frac{\text{v}}{\text{v}+2}=\pm\text{cx}^{-2}$ $\text{Putting}\ \text{v}=\frac{\text{y}}{\text{x}},\ \ \frac{\frac{\text{y}}{\text{x}}}{\frac{\text{y}}{\text{x}}+2}=\pm\text{cx}^{-2}\ \ $ $\Rightarrow\ \ \frac{\text{y}}{\text{y}+2\text{x}}=\pm\text{cx}^{-2}$ $\Rightarrow\ \ \text{x}^{2}\text{y}=\text{C}(\text{y}+2\text{x})\ \ \text{where C}=\pm\text{c}\ \ .....(\text{ii})$ Now putting x = 1 and y = 1 in eq. (ii), we get $1=3\text{C}\ \ \Rightarrow\ \ \text{C}=\frac{1}{3}$
Putting value of C in eq. (ii),
$\text{x}^{2}\text{y}=\frac{1}{3}(\text{y}+2\text{x})\ \ \Rightarrow\ \ 3\text{x}^2\text{y}=\text{y}+2\text{x}$
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