What is a solution to the differential equation dydx=2ex−y with the initial condition y(1)=ln⁡(2e+1)?

Explanation:this is separabledydx=2ex−y=2exe−ySoeydydx=2ex∫eydydxdx=∫2exdx∫ddx(ey)dx=∫2exdxey=2ex+Cy=ln⁡(2ex+C)y(1)=ln⁡(2e+1)ln⁡(2e+1)=ln⁡(2e+C)⇒C=1y=ln⁡(2ex+1)

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High School
Calculus and Analysis
Calculus 2
Applications of integrals

Ansley Sparks

Answered question

2022-04-16

What is a solution to the differential equation

\frac{d y}{d x} = 2 e^{x - y}

with the initial condition

y (1) = \ln (2 e + 1)

Answer & Explanation

phoenixtreeaung

Beginner2022-04-17Added 19 answers

Explanation:
this is separable

\frac{d y}{d x} = 2 e^{x - y} = 2 e^{x} e^{- y}

e^{y} \frac{d y}{d x} = 2 e^{x}

\int e^{y} \frac{d y}{d x} d x = \int 2 e^{x} d x

\int \frac{d}{d x} (e^{y}) d x = \int 2 e^{x} d x

e^{y} = 2 e^{x} + C

y = \ln (2 e^{x} + C)

y (1) = \ln (2 e + 1)

\ln (2 e + 1) = \ln (2 e + C) \Rightarrow C = 1

y = \ln (2 e^{x} + 1)

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