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DIFFERENTIAL EQUATIONS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDIFFERENTIAL EQUATIONS5 Marks
Question
For the differential equaton $\text{xy}\frac{\text{dy}}{\text{dx}}=(\text{x}+2)(\text{y}+2).$ Find the solution curve passing through the point (1, -1).
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Answer

We have,
$\text{xy}\frac{\text{dy}}{\text{dx}}=(\text{x}+2)(\text{y}+2)$
$\Rightarrow\frac{\text{y}}{\text{y}+2}\text{dy}=\frac{(\text{x}+2)}{\text{x}}\text{dx}$
Integrating both sides, we get
$\int\frac{\text{y}}{\text{y}+2}\text{dy}=\int\frac{(\text{x}+2)}{\text{x}}\text{dx}$
$\Rightarrow\int\text{dy}-2\int\frac{1}{\text{y}+2}\text{dy}=\int\text{dx}+2\int\frac{1}{\text{x}}\text{dx}$
$\Rightarrow\text{y}-2\log|\text{y}+2|=\text{x}+2\log|\text{x}|+\text{C}...(1)$
This equation represents the family of solution curves of the given differential equation.
We have to find a particular member of the family, which passes through the point (1, -1).
Substituting $\text{x}=1$ and $\text{y}=-1$ in (1), we get
$-1-2\log|1|=1+2\log|1|+\text{C}$
$\Rightarrow\text{C}=-2$
Putting $\text{C}=-2$ in (1), we get
$\text{y}-2\log|\text{y}+2|=\text{x}+2\log|\text{x}|-2$
$\Rightarrow\text{y}-\text{x}+2=\log\Big\{\text{x}^2(\text{y}+2)^2\Big\}$
Hence, $\text{y}-\text{x}+2=\log\Big\{\text{x}^2(\text{y}+2)^2\Big\}$ is the equation of the required curve.

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