Question
Solve the following differential equations:
$\text{x}^2\text{dy}+\text{y}(\text{x + y})\text{dx}=0$
$\text{x}^2\text{dy}+\text{y}(\text{x + y})\text{dx}=0$
✓
Answer
We have,
$\text{x}^2\text{dy}+\text{y}(\text{x + y})\text{dx}=0$
$\Rightarrow\ \text{x}^2\text{dy}=-\text{y}(\text{x + y})\text{dx}$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{-\text{y}(\text{x + y})}{\text{x}^2}$
This is a homogeneous differential equation.
Putting y = vx and $\frac{\text{dy}}{\text{dx}}=\text{v + x}\frac{\text{dv}}{\text{dx}}$, we get
$\text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{-\text{vx}(\text{x + vx})}{\text{x}^2}$
$\Rightarrow\ \text{v + x}\frac{\text{dv}}{\text{dx}}=-\text{v}(1+\text{v})$
$\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=-\text{v}-\text{v}-\text{v}^2$
$\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=-(\text{v}^2+2\text{v})$
$\Rightarrow\ \frac{\text{dv}}{(\text{v}^2+2\text{v})}=-\frac{\text{dx}}{\text{x}}$
$\Rightarrow\ \frac{\text{dv}}{\text{v}(\text{v}+2)}=-\frac{\text{dx}}{\text{x}}$
Integrating both sides, we get
$\int\frac{\text{dv}}{\text{v}(\text{v}+2)}=-\int\frac{\text{dx}}{\text{x}}$
$\Rightarrow\ \frac{1}2\int\Big[\frac{1}{\text{v}}-\frac{1}{\text{v}+2}\Big]\text{dv}=-\int\frac{\text{dx}}{\text{x}}$
$\Rightarrow\ \frac{1}2\Big[\int\frac{1}{\text{v}}\text{dv}-\int\frac{1}{\text{v}+2}\text{dv}\Big]=-\int\frac{\text{dx}}{\text{x}}$
$\Rightarrow\ \frac{1}2\big[\log|\text{v}|-\log|\text{v}+2|\big]=-\log|\text{x}|+\log\text{C}$
$\Rightarrow\ \frac{1}2\log\Big|\frac{\text{v}}{\text{v}+2}\Big|=\log\Big|\frac{\text{C}}{\text{x}}\Big|$
$\Rightarrow\ \log\Big|\frac{\text{v}}{\text{v}+2}\Big|=\log\Big|\frac{\text{C}^2}{\text{x}^2}\Big|$
$\Rightarrow\ \frac{\text{v}}{\text{v}+2}=\frac{\text{C}^2}{\text{x}^2}$
$\Rightarrow\ \frac{\frac{\text{y}}{\text{x}}}{\frac{\text{y}}{\text{x}}+2}=\frac{\text{C}^2}{\text{x}^2}$
$\Rightarrow\ \frac{\text{y}}{\text{y}+2\text{x}}=\frac{\text{C}^2}{\text{x}^2}$
$\Rightarrow\ \text{x}^2\text{y}=\text{C}^2(\text{y +2x})$
$\Rightarrow\ \text{x}^2\text{y}=\text{K}(\text{y +2x})$ (Where, $K = C^2$)
$\text{x}^2\text{dy}+\text{y}(\text{x + y})\text{dx}=0$
$\Rightarrow\ \text{x}^2\text{dy}=-\text{y}(\text{x + y})\text{dx}$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{-\text{y}(\text{x + y})}{\text{x}^2}$
This is a homogeneous differential equation.
Putting y = vx and $\frac{\text{dy}}{\text{dx}}=\text{v + x}\frac{\text{dv}}{\text{dx}}$, we get
$\text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{-\text{vx}(\text{x + vx})}{\text{x}^2}$
$\Rightarrow\ \text{v + x}\frac{\text{dv}}{\text{dx}}=-\text{v}(1+\text{v})$
$\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=-\text{v}-\text{v}-\text{v}^2$
$\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=-(\text{v}^2+2\text{v})$
$\Rightarrow\ \frac{\text{dv}}{(\text{v}^2+2\text{v})}=-\frac{\text{dx}}{\text{x}}$
$\Rightarrow\ \frac{\text{dv}}{\text{v}(\text{v}+2)}=-\frac{\text{dx}}{\text{x}}$
Integrating both sides, we get
$\int\frac{\text{dv}}{\text{v}(\text{v}+2)}=-\int\frac{\text{dx}}{\text{x}}$
$\Rightarrow\ \frac{1}2\int\Big[\frac{1}{\text{v}}-\frac{1}{\text{v}+2}\Big]\text{dv}=-\int\frac{\text{dx}}{\text{x}}$
$\Rightarrow\ \frac{1}2\Big[\int\frac{1}{\text{v}}\text{dv}-\int\frac{1}{\text{v}+2}\text{dv}\Big]=-\int\frac{\text{dx}}{\text{x}}$
$\Rightarrow\ \frac{1}2\big[\log|\text{v}|-\log|\text{v}+2|\big]=-\log|\text{x}|+\log\text{C}$
$\Rightarrow\ \frac{1}2\log\Big|\frac{\text{v}}{\text{v}+2}\Big|=\log\Big|\frac{\text{C}}{\text{x}}\Big|$
$\Rightarrow\ \log\Big|\frac{\text{v}}{\text{v}+2}\Big|=\log\Big|\frac{\text{C}^2}{\text{x}^2}\Big|$
$\Rightarrow\ \frac{\text{v}}{\text{v}+2}=\frac{\text{C}^2}{\text{x}^2}$
$\Rightarrow\ \frac{\frac{\text{y}}{\text{x}}}{\frac{\text{y}}{\text{x}}+2}=\frac{\text{C}^2}{\text{x}^2}$
$\Rightarrow\ \frac{\text{y}}{\text{y}+2\text{x}}=\frac{\text{C}^2}{\text{x}^2}$
$\Rightarrow\ \text{x}^2\text{y}=\text{C}^2(\text{y +2x})$
$\Rightarrow\ \text{x}^2\text{y}=\text{K}(\text{y +2x})$ (Where, $K = C^2$)
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