How do you solve the differential equation y'=e^(−y)(2x−4), where y5)=0 ?

deiluefniwf

deiluefniwf

Answered question

2022-09-20

How do you solve the differential equation y = e - y ( 2 x - 4 ) , where y5)=0 ?

Answer & Explanation

Ashly Sanford

Ashly Sanford

Beginner2022-09-21Added 9 answers

This is an example of a separable equation with an initial value.
d y d x = e - y ( 2 x - 4 )
y ( 5 ) = 0

by multiplying by e y and by dx,
e y d y = ( 2 x - 4 ) d x
by integrating,
e y d y = ( 2 x - 4 ) d x
e y = x 2 - 4 x + C
by taking the natural log,
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
5 + C = 1 C = - 4

Hence, the solution is y = ln ( x 2 - 4 x - 4 )

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Differential Equations

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?