Integral with 2 variables

[f1player]

dy/dx = 2 - x + y

The question tells us that the integral is equal to: y = 2e^(x) + x - 1 and it wants
us to prove that when you differentiate this, it equals to the above expression
for dy/dx.

But all I get is this: y = 2e^(x) + x - 1

dy/dx = 2e^(x) * 1 + 1
So dy/dx = 2e^(x) + 1 and not 2 - x + y

Can anyone help??

 

[f1player]

So would this be right:

dy/dx = 2 + y - x
= 2 + (2e^x + x - 1) - x
= 2 + (2e^x + 1)
= 1 + 2e^x

Therefore y = x + 2e^x + c

2e^x + x - 1 = x + 2e^x + c

So -1 = c

Therefore y = x + 2e^x - 1 or 2e^x + x - 1

 

[soroban]

Hello, f1player!

Evidently, you're beginning Differential Equations
where they ask you to verify a solution to a differential euqation.

It's like: Verify that \(\displaystyle x\,=\,3\) is a solution of: \(\displaystyle \,2x\,+\,1\:=\:3x\,-\,2\)

\(\displaystyle \frac{dy}{dx}\;=\;2\,-\,x\,+\,y\)

The question tells us that the solution is : \(\displaystyle \,y \:= \;2e^x\,+\,x\,-\,1\)
and it wants us to prove that when you differentiate this,
\(\displaystyle \;\;\)it equals to the above expression for \(\displaystyle \frac{dy}{dx}\)

The left side of that equation is: \(\displaystyle \:\frac{dy}{dx}\) . . . which is: \(\displaystyle \,2e^x\,+\,1\)

The right side is: \(\displaystyle \:2\,-\,x\,+\,y\) . . . which is: \(\displaystyle \,2\,-\,x\,+\,\left(2e^x\,+\,x\,-\,1\right)\;=\;2e^x\,+\,1\)

. . . See?

 

[f1player]

Yes, I see it now

Thanks a lot