A differential equation question

[mech-eng]

There is differential equation with initial condition perplexing me.

y'+ y = 1, y = ce^-x + 1 , y = 2.5 when x = 0

First I take derivative of y which is -ce^-x then I sum it up with y which is -ce^-x+ce^-x + 1 equals 1 which is in harmony with y' + y = 1 but it
seems that this is independent from integral constant c and so there are infinite number of c but in the answer c is -1.5.

Have a nice day.

 

[abitslow]

?
y=1+c*e^(-x); at x= 0, y = 1+c*e° = 1+c=2.5, ...so, c = 1.5

 

[mech-eng]

?
y=1+c*e^(-x); at x= 0, y = 1+c*e° = 1+c=2.5, ...so, c = 1.5

Thank you.

 

[HallsofIvy]

You were told, when they said "y= ce^{-x}+ 1" that this would satisfy the differential equation for all values of c. That is what that means. The problem was to find c such that y(0)= 2.5.

 

[blue_raver22]

I would suggest taking a closer look at the initial condition given. The given initial condition states that y = 2.5 when x = 0. Plugging this into the general solution of y = ce^-x + 1 yields 2.5 = c + 1. This means that c must be equal to 1.5 in order for the initial condition to be satisfied. Therefore, the final solution to the differential equation is y = 1.5e^-x + 1. It is important to carefully consider the initial condition when solving differential equations to ensure that the solution is accurate and complete.