What is a solution to the differential equation exdydx+2exy=1?

exdydx+2exy=1dydx+2y=e−xthis is non-separable, we use an integrating factor (IF)IF=e∫2dx=e2xe2xdydx+2e2xy=e−x⋅e2xe2xdydx+2e2xy=exddxye2x=exye2x=∫exdxye2x=ex+Cy=e−x+Ce−2x

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High School
Calculus and Analysis
Calculus 2
Applications of integrals

Leonard Montes

Answered question

2022-04-19

What is a solution to the differential equation

e^{x} \frac{d y}{d x} + 2 e^{x} y = 1

Answer & Explanation

xxkrnjangxxed9q

Beginner2022-04-20Added 14 answers

e^{x} \frac{d y}{d x} + 2 e^{x} y = 1

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

this is non-separable, we use an integrating factor (IF)

I F = e^{\int 2 d x} = e^{2 x}

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

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

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

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

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

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

shulsellsmine2aaf

Beginner2022-04-21Added 5 answers

Dividing by

e^{x}

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

this linear non-homogeneus differential equation has as solution

y = y_{h} + y_{p}

such that

\frac{{d y}_{h}}{d x} + 2 y_{h} = 0

and

\frac{{d y}_{p}}{d x} + 2 y_{p} = e^{- x}

For the homogeneus we obtain easily ( variables grouping)

y_{h} = C_{0} e^{- 2 x}

and for the particular,

y_{p} = e^{- x}

verifies the differential condition.
so the solution is

y = C_{0} e^{- 2 x} + e^{- x}

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