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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:
$\Big\{\text{x}\sin^2\Big(\frac{\text{y}}{\text{x}}\Big)-\text{y}\Big\}\text{dx + x dy}=0,\text{y}(1)=\frac{\pi}4$
✓

Answer

$\Big\{\text{x}\sin^2\Big(\frac{\text{y}}{\text{x}}\Big)-\text{y}\Big\}\text{dx + x dy}=0,\text{y}(1)=\frac{\pi}4$
It is a homogeneous equation. so, we put y = vx
$\frac{\text{dy}}{\text{dx}}=\text{v + x}\frac{\text{dv}}{\text{dx}}$
So, $\text{v + x}\frac{\text{dv}}{\text{dx}}=-\sin^2\Big(\frac{\text{vx}}{\text{x}}\Big)+\frac{\text{vx}}{\text{x}}$
$\text{x}\frac{\text{dv}}{\text{dx}}=-\sin^2\text{v}$
$\frac{\text{dv}}{\sin^2\text{v}}=-\frac{\text{dx}}{\text{x}}$
integrating both sides, we get
$\cot\Big(\frac{\text{y}}{\text{x}}\Big)=\log|\text{Cx}|$
Putting the values of x = 1 and $\text{y}=\frac{\pi}4$
$\cot\Big(\frac{\pi}{4}\Big)=\log\text{C}$
$1=\log\text{C}$
$\text{C}=\text{e}$
Hence, $\cot\Big(\frac{\text{y}}{\text{x}}\Big)=\log(\text{ex})$

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