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DIFFERENTIAL EQUATIONS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDIFFERENTIAL EQUATIONS4 Marks
Question
For the differential equation $\text{xy}\frac{\text{dy}}{\text{dx}}=(\text{x}+2)(\text{y}+2),$ find the solution curve passing through the point (1, - 1).
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Answer

The given differential equation is

$\text{xy}\frac{\text{dy}}{\text{dx}}=(\text{x}+2)(\text{y}+2),$

Separating the variables, we get,

$\frac{\text{y}}{\text{y}+2}\text{dy}=\frac{\text{x}+2}{\text{x}}\text{dx}$

$\text{Integrating},\ \int\frac{\text{y}}{\text{y}+2}\text{dy}=\int\frac{\text{x}+2}{\text{x}}\text{dx}$

$\therefore\ \int\frac{(\text{y}+2)-2}{\text{y}+2}\text{dy}=\int\bigg[\frac{\text{x}}{\text{x}}+\frac{2}{\text{x}}\bigg]\text{dx}$

$\therefore\ \int\bigg[1-\frac{1}{\text{y}+2}\bigg]\text{dy}=\int\bigg[1+\frac{2}{\text{x}}\bigg]\text{dx}$

$\therefore\ \text{y}-2\log|\text{y}+2|=\text{x}+2\log|\text{x}|+\text{c}\ ...(1)$

Since the curve passes through (1, - 1)

$\therefore\ -1-2\log|-1+2|=1+2\log|1|+\text{c}$

$\therefore\ -1-2\log|1|=1+2\log|1|+\text{c}$

$\therefore\ -1-2(0)=1+2(0)+\text{c}\ \Rightarrow\ \text{c}=-2$

$\therefore\ \text{from}(1),\ \text{y}-2\log|\text{y}+2|=\text{x}+2\log|\text{x}|-2$

$\text{or}\ \text{y}-\text{x}+2=2\log|\text{x}(\text{y}+2)|$

$\text{or}\ \text{y}-\text{x}+2=\log|\text{x}(\text{y}+2)|^2$

$\text{or}\ \text{y}-\text{x}+2=\log[\text{x}^2(\text{y}+2)^2]$

is the required solution.

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