Find the equation of a curve passing through (1,π4) if the slope of the tangent to the curve at any point P(x, y) is yx-cos2 yx.

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`.

बेरीज

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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Forms of Solving Differential Equations> Homogeneous Differential Equations

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Q 9 Q 8 Q 10

पाठ 9: Differential Equations - Solved Examples [पृष्ठ १८४]

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एनसीईआरटी एक्झांप्लर Mathematics Exemplar [English] Class 12

पाठ 9 Differential Equations
Solved Examples | Q 9 | पृष्ठ १८४

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