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प्रश्न
Find the equation of a curve passing through `(1, pi/4)` if the slope of the tangent to the curve at any point P(x, y) is `y/x - cos^2 y/x`.
उत्तर
According to the given condition
`"dy"/"dx" = y/x - cos^2 y/x` .....(i)
This is a homogeneous differential equation.
Substituting y = vx, we get
`"v" + x "dv"/"dx"` = v – cos2v
⇒ `x "dv"/"dx"` = – cos2v
⇒ sec2v dv = `- "dv"/x`
⇒ tan v = – logx + c
⇒ `tan y/x + log x` = c ....(ii)
Substituting x = 1
y = `pi/4`
We get c = 1
Thus, we get `tan (y/x) + log x` = 1, which is the required equation.
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