how to solve this differential equation

[gotit]

the differential equation is

dy/dx + 1 = e^x-y

I have tried this
dy/dx + 1 = e^{x-y ......}equation 1
or, dy/dx=e^x-y-1

taking x-y=v ....equation 2
differentiating eqn 2 with respect to x, we get:
1-dy/dx=dv/dx

or,dy/dx=1-dv/dx

or,e^v-1=1-dv/dx

or,2-e^v=dv/dx

or,dv/2-e^v=dx

i am stuck here. i think this may be simple if i have not mistaken above. can you please help me...

 

[Deleted member 4993]

the differential equation is

dy/dx + 1 = e^x-y

Hint: substitute:

u = e^y


You need to read the rules of this forum. Please read the post titled "Read before Posting" at the following URL:

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We can help - we only help after you have shown your work - or ask a specific question (not a statement like "Don't know any of these")

Please share your work with us indicating exactly where you are stuck - so that we may know where to begin to help you.

 

[Deleted member 4993]

the differential equation is

dy/dx + 1 = e^x-y

I have tried this
dy/dx + 1 = e^{x-y ......}equation 1
or, dy/dx=e^x-y-1

taking x-y=v ....equation 2
differentiating eqn 2 with respect to x, we get:
1-dy/dx=dv/dx

or,dy/dx=1-dv/dx

or,e^v-1=1-dv/dx

or,2-e^v=dv/dx

or,dv/2-e^v=dx

i am stuck here. i think this may be simple if i have not mistaken above. can you please help me...

Try (as suggested in previous post):

u = e^y

It will be much easier to solve!!

 

[HallsofIvy]

the differential equation is

dy/dx + 1 = e^x-y

I have tried this
dy/dx + 1 = e^{x-y ......}equation 1
or, dy/dx=e^x-y-1

taking x-y=v ....equation 2
differentiating eqn 2 with respect to x, we get:
1-dy/dx=dv/dx

or,dy/dx=1-dv/dx

or,e^v-1=1-dv/dx

or,2-e^v=dv/dx

or,dv/2-e^v=dx

i am stuck here. i think this may be simple if i have not mistaken above. can you please help me...

Since you have gotten to this point, let \(\displaystyle u= 2- e^v\). Then \(\displaystyle du= -e^vdv\) and, because \(\displaystyle e^v= 2- u\) this becomes \(\displaystyle du= (u- 2)dv\) and \(\displaystyle \frac{du}{u- 2}= dv\) and you have \(\displaystyle du/(u(u-2))= dx\). And that can be integrated by "partial fractions".

I believe that Subhotosh Khan is suggesting that, going back to the original, you use the fact that \(\displaystyle e^{x-y}= e^xe^{-y}= \frac{e^x}{e^y}\).

 

[Deleted member 4993]

the differential equation is

dy/dx + 1 = e^x-y

Let

u = e^y

du/dx = e^y *dy/dx

1/u * du/dx = dy/dx

dy/dx + 1 = e^x-y → du/dx + u = e^x ← This is a standard 1 st. order linear ODE

du/dx + u = e^x

d/dx (u * e^x) = e^2x

u *e^x = 1/2 * e^2x + C

e^x+y = 1/2 * e^2x + C

y = Ln(1/2 * e^2x + C) - x .... continue as needed