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Solve the Differential Equation: (1 + X2) Dy + 2xydx = Cot Xdx - Mathematics

Question

Sum

Solve the differential equation: (1 + x2) dy + 2xy dx = cot x dx

Solution

The given differential equation is

(1 + x2) dy + 2xy dx = cot x dx

`(d"y")/(d"x") + (2"xy")/(1 + "x"^2) = cot"x"/(1+"x"^2)`

This equation is a linear differential equation of the form:

`dy/dx + py = Q ( "where p" = (2x)/(1 + x^2) and Q = (cot x)/(1 + x^2) )`

`"IF" = e^(int pd"x") = e^(int(2"x")/(1+"x"^2) dx) = e^log(1 + "x"^2) = 1 + x^2`

The general solution of the given differential equation is given by the relation,

y( I.F.) = `int ( "Q" xx "I.F.") dx + C`

⇒ `y(1 + x^2) = int  [ (cot x)/(1+ x^2) (1 + x^2)]dx + C`

⇒ `y(1 + x^2) = int cot x dx + c`

⇒ `y(1 + x^2) = log| sin x | + c`.

  Is there an error in this question or solution?
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