What is the solution of d x d y + 2 y = 1 satisfying y ( 0 ) = 0 ?

d x d y + 2 y = 1 ⇒ d x d y = 1 − 2 y ∫ 1 − 2 y d y = ∫ d x − 2 1 lo g ∣1 − 2 y ∣ = x + C at x = 0 , y = 0 ; − 2 1 lo g 1 = 0 + C ⇒ C = 0 1 − 2 y = e − 2 x ⇒ y = 2 1 − e − 2 x

Mathematics - Differential Equation Question with Solution | TestHub

MathematicsDifferential EquationVariable separableEasy2 minPYQ_2010

MathematicsEasysingle choice

What is the solution of $\frac{d y}{d x} + 2 y = 1$ satisfying $y (0) = 0$ ?

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Solution:

$\frac{d y}{d x} + 2 y = 1 \Rightarrow \frac{d y}{d x} = 1 - 2 y \int \frac{d y}{1 - 2 y} = \int d x - \frac{1}{2} lo g ∣1 - 2 y ∣ = x + C at x = 0, y = 0; - \frac{1}{2} lo g 1 = 0 + C \Rightarrow C = 0 1 - 2 y = e^{- 2 x} \Rightarrow y = \frac{1 - e ^{- 2 x}}{2}$

Stream:BITSATSubject:MathematicsTopic:Differential EquationSubtopic:Variable separable

⏱ 2mℹ️ Source: PYQ_2010

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