Solution of the equation d y d x = e x - y e x - e y is

We have, d y d x = e x - y e x - e y ⇒ e y d y d x + e y e x = e 2 x Put e y = v ∴ e y d y d x = d v d x ∴ d v d x + v e x = e 2 x This is linear equation I . F = = e ∫ e x d x = e e x ∴ Required solution is v · e e x = ∫ e e x · e 2 x d x + C v · e e x = e e x e x - 1 + C ⇒ e y · e e x = e e x e x - 1 + C ⇒ e y = e x - 1 + c e - e x

Mathematics - Differential Equation Question with Solution | TestHub

MathematicsDifferential EquationLinear DE / Red. LDEMedium2 minPYQ_2019

MathematicsMediumsingle choice

Solution of the equation $\frac{d y}{d x} = e^{x - y} (e^{x} - e^{y})$ is

Options:

Answer:

Solution:

We have,

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

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

Put $e^{y} = v$

$∴ e^{y} \frac{d y}{d x} = \frac{d v}{d x}$

$∴ \frac{d v}{d x} + v e^{x} = e^{2 x}$

This is linear equation

$I . F = = e^{\int e^{x} d x} = e^{e^{x}}$

$∴$ Required solution is

$v \cdot e^{e^{x}} = \int e^{e^{x}} \cdot e^{2 x} d x + C$

$v \cdot e^{e^{x}} = e^{e^{x}} (e^{x} - 1) + C$

$\Rightarrow e^{y} \cdot e^{e^{x}} = e^{e^{x}} (e^{x} - 1) + C$

$\Rightarrow e^{y} = e^{x} - 1 + c e^{- e^{x}}$

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

⏱ 2mℹ️ Source: PYQ_2019

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