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प्रश्न
Solve the following differential equation:
`xdy/dx=y-xsin^2(y/x)`, given that y(1) = `pi/6`
योग
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उत्तर
`xdy/dx=y-xsin^2(y/x)`
`dy/dx=y/x-sin^2(y/x)`
Which is a homogeneous diff. equation
y = vx
⇒ `dy/dx=v+x(dv)/dx`
⇒ `v+x(dv)/dx=(vx)/x-sin^2((vx)/x)`
⇒ `v+x(dv)/dx=v-sin^2v`
⇒ `(dv)/sin^2v=-dx/x`
integrating both sides,
⇒ `intcosec^2vdv=-intdx/x`
⇒ –cotv = –log|x| + C
⇒ cot v = log|x| – C
Put v = `y/x`
cot `(y/x)` = log|x| – C
y(1) = `pi/6`; x = 1, y = `pi/6`
Put, cot `(pi)/6=log|1| -C` ...[∵ log|1| = 0]
⇒ `sqrt3=-C`
⇒ C = `-sqrt3`
`cot(y/x)=log|x|+sqrt3`
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