Get the answer to "How do you solve the differential equation y′=e−y(2x−4),..."

Ashleigh Mitchell

Answered question

2022-04-11

How do you solve the differential equation

y^{'} = e^{- y} (2 x - 4)

, where y(5)=0?

razmenile7chp

Beginner2022-04-12Added 11 answers

This is an example of a separable equation with an initial value.

\frac{d y}{d x} = e^{- y} (2 x - 4)

y(5)=0
by multiplying by

e^{y}

and by dx,

\Rightarrow e^{y} d y = (2 x - 4) d x

by integrating,

\Rightarrow \int e^{y} d y = \int (2 x - 4) d x

\Rightarrow e^{y} = x^{2} - 4 x + C

by taking the natural log,

\Rightarrow y = \ln (x^{2} - 4 x + C)

Now, we need to find C using y(5)=0,

y (5) = \ln ({(5)}^{2} - 4 (5) + C) = \ln (5 + C) = 0

\Rightarrow 5 + C = 1 \Rightarrow C = - 4

Hence, the solution is

y = \ln (x^{2} - 4 x - 4)

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