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MARCH 2024 — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsMARCH 20243 Marks
Question
Solve the following differential equation:
$x \frac{d y}{d x}-y+x \sin \left(\frac{y}{x}\right)=0$
✓

Answer

$ x \frac{d y}{d x}-y+x \sin \left(\frac{y}{x}\right)=0 $ .....(1)
Put y = vx
$\therefore \frac{ dy }{ dx }= V + x \frac{ dV }{ dx }$ and $\frac{ y }{ x }= V$
$\therefore$ (1) becomes, $x \left( V + x \frac{ dV }{ dx }\right)- Vx + x \sin V=0$
$\therefore V x+x^2 \frac{d V}{d x}-V x+x \sin V=0$
$\therefore x ^2 \frac{ dV }{ dx }+ x \sin V =0$
$\therefore \frac{1}{\sin V} d V+\frac{1}{x} d x=0$
Integrating, we get
$\therefore \int \operatorname{cosec} VdV +\int \frac{1}{ x } dx = c _1$
$\therefore \log |\operatorname{cosec} V -\cot V |+\log | x |=\log c$, where $c _1=\log c$
$\therefore \log | x (\operatorname{cosec} V -\cot V )|=\log c$
$\therefore x\left(\frac{1}{\sin V}-\frac{\cos V}{\sin V}\right)=c$
$\therefore x(1-\cos V)=c \sin V$
$\therefore x\left[1-\cos \left(\frac{y}{x}\right)\right]=c \sin \left(\frac{y}{x}\right)$
This is the general solution.

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