Let y - yx be the solution of the differential equation dydx - 2y-2sin x - 5x-2cos x such that y0 - 7- Then y- is equal to

Given: dydx = 2(y+2sin x 5)x 2cos x ⇒dydx 2xy = 4xsin x 2cos x 10x I.F. = e^∫ 2xdx = e^ x^2 Now, the general solution: y· e^ x^2=∫ e^ x^2(4xsin x 2cos x 10x ...

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Standard XII

Mathematics

Order of a Differential Equation

Let y = yx be...

Question

Let $y = y (x)$ be the solution of the differential equation $\frac{d y}{d x} = 2 (y + 2 sin x - 5) x - 2 cos x$ such that $y (0) = 7$ . Then $y (π)$ is equal to

e^{π^{2}} + 5

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2 e^{π^{2}} + 5

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7 e^{π^{2}} + 5

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3 e^{π^{2}} + 5

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Solution

The correct option is B $2 e^{π^{2}} + 5$
Given: $\frac{d y}{d x} = 2 (y + 2 sin x - 5) x - 2 cos x$
$\Rightarrow \frac{d y}{d x} - 2 x y = 4 x sin x - 2 cos x - 10 x$
$I . F . = e^{\int - 2 x d x} = e^{- x^{2}}$
Now, the general solution:
$y \cdot e^{- x^{2}} = \int e^{- x^{2}} (4 x sin x - 2 cos x - 10 x) d x + C$
$\Rightarrow y e^{- x^{2}} = 4 \int e^{- x^{2}} (x) (sin x) d x - 2 \int (cos x) e^{- x^{2}} d x + 5 \int (- 2 x e^{- x^{2}}) d x + C$
$\Rightarrow y e^{- x^{2}} = - 2 sin x \cdot e^{- x^{2}} + 5 e^{- x^{2}} + C$
Put $x = 0 \Rightarrow 7 = 0 + 5 + C$
$\Rightarrow C = 2$
$∴ y e^{- x^{2}} = - 2 sin x \cdot e^{- x^{2}} + 5 e^{- x^{2}} + 2$
$\Rightarrow y = - 2 sin x + 5 + 2 e^{x^{2}}$
Put $x = π \Rightarrow y = 5 + 2 e^{π^{2}}$

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Let y=y(x) be the solution of the differential equation dydx=2(y+2sinx−5)x−2cosx such that y(0)=7. Then y(π) is equal to

Let $y = y (x)$ be the solution of the differential equation $\frac{d y}{d x} = 2 (y + 2 sin x - 5) x - 2 cos x$ such that $y (0) = 7$ . Then $y (π)$ is equal to