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प्रश्न
The solution of the differential equation `("d"y)/("d"x) + (2xy)/(1 + x^2) = 1/(1 + x^2)^2` is ______.
पर्याय
y(1 + x2) = c + tan–1x
`y/(1 + x^2) = "c" + tan^-1x`
y log(1 + x2) = c + tan–1x
y(1 + x2) = c + sin–1x
उत्तर
The solution of the differential equation `("d"y)/("d"x) + (2xy)/(1 + x^2) = 1/(1 + x^2)^2` is y(1 + x2) = c + tan–1x.
Explanation:
The given differential equation is `("d"y)/("d"x) + (2xy)/(1 + x^2) = 1/(1 + x^2)^2`
Since, it is a linear differential equation
P = `(2x)/(1 + x^2)` and Q = `1/(1 + x^2)^2`
Integrating factor I.F. = `"e"^(int Pdx)`
= `"e"^(int (2x)/(1 + x^2) "d"x)`
= `"e"^(log(1 + x^2))`
= `(1 + x^2)`
∴ Solution is `y xx "I"."F". = int "Q" xx "I"."F". "d"x + "c"`
⇒ `y(1 + x^2) = int 1/(1 + x^2)^2 xx (1 + x^2)"d"x + "c"`
⇒ `y(1 + x^2) = int 1/((1 + x^2)) "d"x + "c"`
⇒ `y(1 + x^2) = tan^-1x + "c"`.
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