हिंदी

For the given below, verify that the given function (implicit or explicit) is a solution to the corresponding differential equation. y=ex(acosx+bsinx) : d2ydx2-2dydx+2y=0

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प्रश्न

For the given below, verify that the given function (implicit or explicit) is a solution to the corresponding differential equation.

`y = e^x (acos x + b sin x)  :  (d^2y)/(dx^2) - 2 dy/dx + 2y = 0`

योग
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उत्तर

Given function y = ex (a cos x + b sin x)

On differentiating

`dy/dx = e^x (a sin x + b cos x) + e^x (a cos x + b sin x)`

`= e^x [(a + b) cos x + (b - a) sin x]`

On differentiating again,

`(d^2y)/dx^2 = e^x [- (a + b) sin x + (b - a) cos x] + e^x [(a + b) cos x + (b - a) sin x]`

`(d^2y)/(dx^2)= e^x [2b cos x - 2a sin x]`

L.H.S. ⇒ `(d^2y)/(dx^2) - 2 dy/dx + 2y`

`= e^x [2b cos x - 2a sin x] - 2e^x [(a + b) cos x + (b - a) sin x] + 2e^x [a cos x + b sin x] = 0`

`= e^x [(2b - 2a - 2b + 2a)] cos x + (- 2a - 2b + 2a + 2b) sin x`

`= "e"^x xx 0 + 0 xx sin x = 0`  R.H.S.

Hence the given function is a solution to the differential equation.

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अध्याय 9: Differential Equations - Exercise 9.7 [पृष्ठ ४२०]

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एनसीईआरटी Mathematics Part 1 and 2 [English] Class 12
अध्याय 9 Differential Equations
Exercise 9.7 | Q 2.2 | पृष्ठ ४२०

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