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Solve the following differential equation: ddxdydx=y-xcos2(yx)

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प्रश्न

Solve the following differential equation:

`x ("d"y)/("d"x) = y - xcos^2(y/x)`

बेरीज
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उत्तर

Given `x ("d"y)/("d"x) = y - x cos^2 y/x`

The equation can be written as

`("d"y)/("d"x) = (y - cos^2 y/x)/x` ..........(1)

This is a homogeneous differential equation.

y = vx

`("d"y)/("d"x) = "v"(1) + x "dv"/("d"x)`

Substituting `("d"y)/("d"x)` value in equation (1), we get

`"v" + (x"dv")/("d"x) = ("v"x - x cos^2 ((vx)/x))/x`

`"v" + (x"dv")/("d"x) = ("v"x - x cos^2("v"))/x`

`"v" + (x"dv")/("d"x) = x (("v" - cos^2"v"))/x`

`x "dv"/("d"x) = "v" - cos^2"v" - "v"`

`"dv"/("d"x) = (- cos^2"v")/x`

`"dv"/(cos^2"v") = (-"d"x)/x`

Integrating on both sides, we get

`int sec^2"v"  "d"x = - int ("d"x)/x`

tan v = – log x + log C

tan v = log C – log x

tan v = `log ("C"/x)`

etan v = `"C"/x`

C = xetan v  

C = `xe"^(tan  y/x)` is a required equation.

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Solution of First Order and First Degree Differential Equations
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 10: Ordinary Differential Equations - Exercise 10.6 [पृष्ठ १६६]

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सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 12 TN Board
पाठ 10 Ordinary Differential Equations
Exercise 10.6 | Q 6 | पृष्ठ १६६

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