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Solve : dydxx2dydx = x2 + xy + y2.

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प्रश्न

Solve : `x^2 "dy"/"dx"` = x2 + xy + y2.

योग
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उत्तर

Given equation is `x^2 "dy"/"dx"` = x2 + xy + y2 

⇒ `"dy"/"dx" = (x^2 + xy + y^2)/x^2`

Put y = vx  ......[∵ it is a homogeneous differential equation]

∴ `"dy"/"dx" = "v" + x * "dv"/"dx"`

∴ `"v" + x * "dv"/"dx" = (x^2 + "v"x^2 + "v"^2x^2)/x^2`

⇒ `"v" + x * "dv"/"dx" = (x^2(1 + "v" + v"^2))/x^2` 

⇒ `"v" + x * "dv"/"dx" = 1 + "v" + "v"^2`

⇒ `x * "dv"/"dx" = 1 + "v" + "v"^2 -  "v"`

⇒ `x * "dv"/"dx" = 1 + "v"^2`

⇒ `"dv"/(1 + "v"^2) = "dx"/x`

Integrating both sides, we get

`int "dv"/(1 + "v"^2) = int "dx"/x`

⇒ tan–1v = log x + c

⇒ `tan^-1 (y/x)` = log x + c

Hence, the required solution is `tan^-1 (y/x)` = log |x| + c.

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अध्याय 9: Differential Equations - Exercise [पृष्ठ १९४]

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एनसीईआरटी एक्झांप्लर Mathematics Exemplar [English] Class 12
अध्याय 9 Differential Equations
Exercise | Q 16 | पृष्ठ १९४

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