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For the differential equation, find the general solution: xdydx + 2y=x2logx - Mathematics

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प्रश्न

For the differential equation, find the general solution:

`x dy/dx +  2y= x^2 log x`

योग

उत्तर

The given equation

`x dy/dx + 2y = x^2 log x`

or `dy/dx + (2/x)y = x log x`

Comparing with `dy/dx + Py = Q`,

P = `2/x` and Q = x log x

∴ I.F. = `e^(int P dx) = e^(int_x^2 dx)`

`= e^(2 log x) = e^(log x^2) = x^2`

Hence the required solution

∴ y × I.F. = ∫ Q × I.F. dx + C

⇒ y × x2 = ∫ x2 + x log x dx + C

⇒ x2 y = ∫ x3 log x + C

⇒ x2 y = `log x * x^4/4 - int 1/4 * x^4/4 dx + C`

⇒ x2 y = `x^4/4 log x - 1/4 int x^3 dx + C`

⇒ x2 y = `x^4/4 log x - 1/4 xx x^4/4 + C`

⇒ y = `x^2/16 (4 log x - 1) + C/x^2`

Which is the required solution.

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अध्याय 9: Differential Equations - Exercise 9.6 [पृष्ठ ४१३]

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एनसीईआरटी Mathematics [English] Class 12
अध्याय 9 Differential Equations
Exercise 9.6 | Q 6 | पृष्ठ ४१३

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