Solve the following equation: x dy dx +y=y^2 - Vidyadip

Question

Solve the following equation:
$\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}=\text{y}^2$

✓

Answer

We have,
$\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}=\text{y}^2$
$\Rightarrow\text{x}\frac{\text{dy}}{\text{dx}}=\text{y}^2-\text{y}$
$\Rightarrow\frac{1}{\text{y}^2-\text{y}}\ \text{dy}=\frac{1}{\text{x}}\ \text{dx}$
integrating both sides, we get
$\int\frac{1}{\text{y}^2-\text{y}}\text{dy}=\int\frac{1}{\text{x}}\text{dx}$
$\Rightarrow\int\frac{1}{\text{y}(\text{y}-1)}=\text{dy}=\int\frac{1}{\text{x}}\text{dx}\ ...(1)$
Let $\frac{1}{(\text{y}-1)}=\frac{\text{A}}{\text{y}}+\frac{\text{B}}{\text{y}-1}$
$\Rightarrow1=\text{A}(\text{y}-1)+\text{B}(\text{y})$
putting y = 0, we get
1 = -A
⇒ A = -1
putting y = 1, we get
1 = B
$\therefore\frac{1}{\text{y}(\text{y}-1)}=\frac{-1}{\text{y}}+\frac{1}{\text{y}-1}$
$\Rightarrow\int\frac{1}{\text{y}(\text{y}-1)}\text{dy}=\int\frac{-1}{\text{y}}\text{dy}+\int\frac{1}{\text{y}-1}\text{dy}\ ...(2)$
From (1) & (2) , we get
$\int\frac{-1}{\text{y}}\text{dy}+\int\frac{1}{\text{y}-1}\text{dy}=\int\frac{1}{\text{x}}\text{dx}$
$\Rightarrow-\log|\text{y}|+\log|\text{y}-1|=\log|\text{x}|+\log\text{C}$
$\Rightarrow\log\Big|\frac{\text{y}-1}{\text{y}}\Big|-\log|\text{x}|=\log\text{C}$
$\Rightarrow\log\Big|\frac{\text{y}-1}{\text{xy}}\Big|=\log\text{C}$
$\Rightarrow\frac{\text{y}-1}{\text{xy}}=\text{C}$
$\Rightarrow\text{y}-1=\text{Cxy}$
hence, $\text{y}-1=\text{Cxy}$ is the required solution.

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