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DIFFERENTIAL EQUATIONS — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSDIFFERENTIAL EQUATIONS5 Marks
Question
Solve the following initial value problems:
$\text{x}(\text{x}^2+3\text{y}^2)\text{dx}+\text{y}(\text{y}^2+3\text{x}^2)\text{dy}=0,\text{y}(1)=1$
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Answer

$\text{x}(\text{x}^2+3\text{y}^2)\text{dx}+\text{y}(\text{y}^2+3\text{x}^2)\text{dy}=0,\text{y}(1)=1$
$\frac{\text{dy}}{\text{dx}}=-\frac{\text{x}(\text{x}^2+3\text{y}^2)}{\text{y}(\text{y}^2+3\text{x}^2)}$
It is a homogeneous equation
put y = vx
$\frac{\text{dy}}{\text{dx}}=\text{v + x}\frac{\text{dv}}{\text{dx}}$
So,
$\text{v + x}\frac{\text{dv}}{\text{dx}}=-\frac{\text{x}(\text{x}^2+3\text{v}^2\text{x}^2)}{\text{vx}(\text{v}^2\text{x}^2+3\text{x}^2)}$
$\text{x}\frac{\text{dv}}{\text{dx}}=\frac{(1+3\text{v}^2)}{\text{v}(\text{v}^2+3)}-\text{v}$
$\text{x}\frac{\text{dv}}{\text{dx}}=\frac{-1-3\text{v}^2-\text{v}^4-3\text{v}^2}{\text{v}(\text{v}^2+3)}$
$=\frac{-\text{v}^4-6\text{v}^2-1}{\text{v}(\text{v}^2+3)}$
$\frac{\text{v}(\text{v}^2+3)}{\text{v}^4+6\text{v}^2+1}\text{dv}=-\frac{\text{dx}}{\text{x}}$
$\int\frac{4\text{v}^3+12\text{v}}{\text{v}^4+6\text{v}^2+1}\text{dv}=-4\int\frac{\text{dx}}{\text{x}}$
$\log|\text{v}^4+6\text{v}^2+1|=\log\Big|\frac{\text{C}}{\text{x}^4}\Big|$
$|\text{v}^4+6\text{v}^2+1|=\Big|\frac{\text{C}}{\text{x}^4}\Big|\ \dots(\text{i})$
Put y = 1, x = 1
(1+6+1) = C
⇒ C = 8
Put C = 8 in equation (i),
$(\text{y}^4+\text{x}^4+6\text{x}^2\text{y}^2)=8$

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