# First order nonlinear ordinary diff eqn

• Nov 28th 2010, 07:11 AM
MattWT
First order nonlinear ordinary diff eqn
Hi guys,

Have the equation

$\displaystyle y' = 1 + y, y(0) = 0$

Show that

$\displaystyle y(t) = e^t - 1$

I've tried a couple of methods, but cannot get to that solution. Maybe my algebra is a little of, unless I got it totally wrong, thanks again.

$\displaystyle \int y' = \int 1 + y dt$
• Nov 28th 2010, 07:20 AM
TheEmptySet
Quote:

Originally Posted by MattWT
Hi guys,

Have the equation

$\displaystyle y' = 1 + y, y(0) = 0$

Show that

$\displaystyle y(t) = e^t - 1$

I've tried a couple of methods, but cannot get to that solution. Maybe my algebra is a little of, unless I got it totally wrong, thanks again.

$\displaystyle \int y' = \int 1 + y dt$

The equation is both linear and seperable

$\displaystyle \displaystyle \frac{dy}{dt}=y+1 \iff \frac{dy}{y+1}=dt$

Now just integrate and use the initial condition.
• Nov 28th 2010, 07:22 AM
harish21
$\displaystyle y' = 1 + y$

$\displaystyle \dfrac{dy}{dt} = 1+y$

$\displaystyle \displaystyle \int \dfrac{1}{1+y}\;dy=\int \;dt$

$\displaystyle \ln (1+y) = t + C$
• Nov 28th 2010, 07:32 AM
HallsofIvy
But apparently this problem does not ask you to solve the problem, only to verify that the given function satisfies the conditions. That is much easier than solving. (For example, it would be difficult to solve the equation $\displaystyle x^9- 3x^3+ 4x- 2= 0$ but very easy to verify that x= 1 is a solution.)

Given that $\displaystyle y= e^x- 1$ it is easy to verify that $\displaystyle y(0)= e^0- 1= 1- 1= 0$. Its derivative is $\displaystyle y'= e^x$ and it is easy to verify that $\displaystyle y'= e^x= 1+ (e^x- 1)= 1+ y$.
• Nov 28th 2010, 10:45 AM
Krizalid
that's something that we discuss with my people too.

since it says "show," to me at least it's like "prove," so in order to do what you mean, it should be "verify."