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If x = x(y) is the solution of the differential equation $y \frac{d x}{d y} = 2 x + y^{3} (y + 1) e^{y}, x (1) = 0$ ; then x(e) is equal to :

Option 1 -

$e^{3} (e^{e} - 1)$

Option 2 -

$e^{e} (e^{3} - 1)$

Option 3 -

$e^{2} (e^{e} + 1)$

Option 4 -

$e^{e} (e^{2} - 1)$

2 Views | Posted 2 months ago

Asked by Shiksha User

1 Answer

Answered by

alok kumar singh | Contributor-Level 10

2 months ago

Correct Option - 1

Detailed Solution:

$y \frac{d x}{d y} = 2 x + y^{3} (y + 1) e^{y}$

$\frac{d x}{d y} - \frac{2}{y} x = y^{3} (1 + y) e^{y} . . . . . . . . . . (i)$

Equation (i) is in linear form, so I.F. = y^-2

$\begin{array}{l} ∴ x y^{- 2} = y e^{y} + c \Rightarrow 0 = e + c \Rightarrow c = - e \\ ∴ x = e^{3} (e^{e} - 1) \end{array}$

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If x = x(y) is the solution of the differential equation $y \frac{d x}{d y} = 2 x + y^{3} (y + 1) e^{y}, x (1) = 0$ ; then x(e) is equal to :

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If x = x(y) is the solution of the differential equation y d x d y = 2 x + y 3 ( y + 1 ) e y , x ( 1 ) = 0 ; then x(e) is equal to :

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If x = x(y) is the solution of the differential equation $y \frac{d x}{d y} = 2 x + y^{3} (y + 1) e^{y}, x (1) = 0$ ; then x(e) is equal to :