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प्रश्न
Find the equation of a curve passing through origin and satisfying the differential equation `(1 + x^2) "dy"/"dx" + 2xy` = 4x2
उत्तर
Given equation is `(1 + x^2) "dy"/"dx" + 2xy` = 4x2
⇒ `"dy"/"dx" + (2x)/(1 + x^2) * y = (4x^2)/(1 + x^2)`
Here, P = `(2x)/(1 + x^2)` and Q = `(4x^2)/(1 + x^2)`
Integrating factor I.F. = `"e"^(int Pdx)`
= `"e"^(int (2x)/(1 + x^2) dx)`
= `"e"^(log(1 + x^2)`
= 1 + x2
∴ Solution is `y xx "I"."F". = int "Q" xx "I"."F". "d"x + "c"`
⇒ `y(1 + x^2) = int (4x^2)/(1 + x^2) xx (1 + x^2) "d"x + "c"`
⇒ `y(1 + x^2) = int 4x^2 "d"x + "c"`
⇒ `y(1 + x^2) = 4/3 x^3 + "c"` ......(i)
Since the curve is passing through origin i.e., (0, 0)
∴ Put y = 0 and x = 0 in equation (i)
0(1 + 0) = `4/3(0)^3 + "c"`
⇒ C = 0
∴ Equation is `y(1 + x^2) = 4/3 x^3`
⇒ y = `(4x^3)/(3(1 + x^2))`
Hence, the required solution is y = `(4x^3)/(3(1 + x^2))`.
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