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प्रश्न
Solve the differential equation: (1 +x2 ) dy + 2xy dx = cot x dx
उत्तर
(1 +x2 ) dy + 2xy dx = cot x dx
(1 +x2 ) `dy/dx ` + 2xy = cot x
`dy/dx +(2xy)/(1+x^2) = (cot x)/(1 +x^2)`
`dy/dx + py = q`
Comparing with linear differential equation
I .F . = `e^(intp dx)`
I.F = `e^(int x/(1+x^2) dx)`
I .F = `int (2x)/(1+x^2) dx`
put t = 1 +x2
`dt/dx = 2x`
dt = 2x dx
`I = int dt/t = "In " t = "In" ( 1+ x^2 )`
`⇒ I .f = e^("in"^((1+x^2) ) = 1 +x^2`
`x(I.F) = int Q ( I .F) dx + c`
`x(1 +x^2 ) = int (cot x ) /(1 +x^2 ) (1+ x^2 ) dx + c `
` x(1 + x^2 ) = int cot x dx + c`
x + x3 = In | sin x | + c
x + x3 - In | sin x | + c + 0
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