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Differential Equations — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDifferential Equations5 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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