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

Given d x d y - 2 x y = y 2 y + 1 e y Finding integrating factor I . F . = e ∫ − 2 y d x = e − 2 ln y = 1 y 2 ⇒ x y 2 = ∫ y + 1 e y d y ⇒ x y 2 = y · e y + C 0 = e + C ⇒ C = - e ⇒ x y 2 = y e y - e Putting y = e in the equation we get ⇒ x e 2 = e . e e - e ⇒ x = e 3 e e - 1

Mathematics - Differential Equation Question with Solution | TestHub

MathematicsDifferential EquationLinear DE / Red. LDEMedium2 minPYQ_2022

MathematicsMediumnumerical range

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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Answer:

Solution:

Given

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

Finding integrating factor $I . F . = e^{\int \frac{- 2}{y} d x} = e^{- 2 \ln y} = \frac{1}{y^{2}}$

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

$\Rightarrow \frac{x}{y^{2}} = y \cdot e^{y} + C$

$0 = e + C \Rightarrow C = - e$

$\Rightarrow \frac{x}{y^{2}} = y e^{y} - e$

Putting $y = e$ in the equation we get

$\Rightarrow \frac{x}{e^{2}} = e . e^{e} - e$

$\Rightarrow x = e^{3} (e^{e} - 1)$

Stream:JEESubject:MathematicsTopic:Differential EquationSubtopic:Linear DE / Red. LDE

⏱ 2mℹ️ Source: PYQ_2022

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