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Differential Equations — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDifferential Equations5 Marks
Question
Find the particular solution of the differential equation$(1-\text{y}^2)(1+\log\text{x})\text{dx}+2\text{xy dy}=0,$ given that $\text{y}=0$ when $\text{x}=1.$
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Answer

Given:
$(1-\text{y}^2)(1+\log\text{x})\text{dx}+2\text{xy dy}=0$
$\Rightarrow(1-\text{y}^2)(1+\log\text{x})\text{dx}=-2\text{x y dy}$
$\Rightarrow\Big(\frac{1+\log\text{x}}{2\text{x}}\Big)\text{dx}=-\Big(\frac{\text{y}}{1-\text{y}^2}\Big)\text{dy}...(1)$
Let:
$1+\log\text{x = t}$
and
$(1-\text{y}^2)=\text{p}$
$\Rightarrow\frac{1}{\text{x}}\text{dx dt}$ and $-2\text{y dy = dp}$
Therefore, (1) becomes
$\int\frac{\text{t}}{2}\text{dt}=\int\frac{1}{2\text{p}}\text{dp}$
$\Rightarrow\frac{\text{t}^2}{4}=\frac{\log\text{p}}{2}+\text{C}...(2)$
Substituting the values of t and p in (2) we get
$\frac{(1+\log\text{x})^2}{4}=\frac{\log(1-\text{y}^2)}{2}+\text{C}...(3)$
At $\text{x}=1$ and $\text{y}=0,$ (3) becomes
$\text{C}=\frac{1}{4}$
Substituting the value of C in (3), we get
$\frac{(1+\log\text{x})^2}{4}=\frac{\log(1-\text{y}^2)}{2}+\frac{1}{4}$
$\Rightarrow(1+\log\text{x})^2=2\log(1-\text{y}^2)+1$
Or
$(\log\text{x})^2+\log\text{x}^2=\log(1-\text{y}^2)^2$
It is the required particular solution.

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