# verify that the pair of functions is a solution...

• Jan 27th 2011, 05:35 PM
slapmaxwell1
verify that the pair of functions is a solution...
verify that the pair of functions x = t + e^t , y = te^t is a solution to the system of differential equations.

x + y - dy/dt = 1 , x - dx/dt = t - 1.

ok so i was able to prove that the pair of functions were solutions to the x + y - dy/dt = 1.

but i couldnt prove they were solutions to the equation on the right, x - dx/dt = t - 1.

• Jan 27th 2011, 05:44 PM
slapmaxwell1
to do this problem i substituted the two equations into the x + y - dy/dt = 1. and solved it for dy/dt, the problem checked out. but i couldnt do that with the other side?? any thoughts?
• Jan 27th 2011, 05:45 PM
Prove It
$\displaystyle x = t + e^t$, so $\displaystyle \frac{dx}{dt} = 1 + e^t$.

What does $\displaystyle x - \frac{dx}{dt}$ equal?
• Jan 27th 2011, 05:47 PM
slapmaxwell1
sorry, X - dx/dt = t - 1
• Jan 27th 2011, 05:50 PM
Prove It
Exactly...

$\displaystyle x - \frac{dx}{dt} = t + e^t - (1 + e^t) = t - 1$.

Therefore we have verified $\displaystyle x = t + e^t$ is a solution to that DE.
• Jan 27th 2011, 05:53 PM
slapmaxwell1
ok but what about the function with the y in it? wouldnt i have to prove that y=te^t is also a solution of x-dx/dt = t-1 ?
• Jan 27th 2011, 06:03 PM
Prove It
How can it be? There's no $\displaystyle y$ in it...
• Jan 27th 2011, 06:06 PM
slapmaxwell1
thats what i was hoping. my prof wanted me to prove that both were solutions. thank you!
• Jan 27th 2011, 06:10 PM
Prove It
Quote:

Originally Posted by slapmaxwell1
thats what i was hoping. my prof wanted me to prove that both were solutions. thank you!

What I mean is, in the second DE $\displaystyle y$ can be anything, as the DE doesn't depend on it.

So to have a solution to both DEs, you just need a $\displaystyle y$ that satisfies the first...