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DIFFERENTIAL EQUATIONS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDIFFERENTIAL EQUATIONS4 Marks
Question
Solve the following initial value problems:
$(\text{x}^2+\text{y}^2)\text{dx}=2\text{xy dy, y}(1)=0$
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Answer

$(\text{x}^2+\text{y}^2)\text{dx}=2\text{xy dy, y}(1)=0$

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

It is a homogeneous equation.

Put y = vx
and $\frac{\text{dy}}{\text{dx}}=\text{v + x}\frac{\text{dv}}{\text{dx}},$

So,

$\text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{\text{x}^2+\text{v}^2\text{x}^2}{2\text{xvx}}$

$\text{x}\frac{\text{dv}}{\text{dx}}=\frac{1+\text{v}^2}{2\text{v}}-\text{v}$

$\text{x}\frac{\text{dv}}{\text{dx}}=\frac{1+\text{v}^2-2\text{v}^2}{2\text{v}}$

$\text{x}\frac{\text{dv}}{\text{dx}}=\frac{1-\text{v}^2}{2\text{v}}$

$\int\frac{2\text{v}}{1-\text{v}^2}=\int\frac{\text{dx}}{\text{x}}$

$\log|1-\text{v}^2|=-\log|\text{x}|+\log|\text{C}|$

$\log|1-\text{v}^2|=\log\Big|\frac{\text{C}}{\text{x}}\Big|$

$\Big|\frac{\text{x}^2-\text{y}^2}{\text{x}^2}\Big|=\Big|\frac{\text{C}}{\text{x}}\Big|$

$|\text{x}^2-\text{y}^2|=|\text{Cx}|\ \dots(\text{i})$

Put y = 0, x = 1

1 - 0 = C

C = 1

Put the value of C in equation (i),

$|\text{x}^2-\text{y}^2|=|\text{x}|$

$(\text{x}^2-\text{y}^2)^2=\text{x}^2$

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