# Thread: Basic diff equations problem

1. ## Basic diff equations problem

Consider $\displaystyle \dot x = x+1$. Find its solution x(t) satisfying the initial condition x(0)=0.

How should I attack this problem?

2. Originally Posted by garymarkhov
Consider $\displaystyle \dot x = x+1$. Find its solution x(t) satisfying the initial condition x(0)=0.

How should I attack this problem?

Remember that $\displaystyle \dot x =\frac{dx}{dt}$

This gives a first order seperable ODE.

$\displaystyle \frac{dx}{dt}=x+1 \iff \frac{dx}{x+1}=dt$

Integrating both sides we get

$\displaystyle \ln|x+1|=t+c$ solving for x gives

$\displaystyle x+1=Ae^{t} \iff x(t)=Ae^{t}-1$

From here just plug in your intial condition to find A.

Good luck.

3. Originally Posted by TheEmptySet
Remember that $\displaystyle \dot x =\frac{dx}{dt}$

This gives a first order seperable ODE.

$\displaystyle \frac{dx}{dt}=x+1 \iff \frac{dx}{x+1}=dt$

Integrating both sides we get

$\displaystyle \ln|x+1|=t+c$ solving for x gives

$\displaystyle x+1=Ae^{t} \iff x(t)=Ae^{t}-1$

From here just plug in your intial condition to find A.

Good luck.
Cool, so A = 1 and $\displaystyle x(t)=e^t$ right?

4. Originally Posted by garymarkhov
Cool, so A = 1 and $\displaystyle x(t)=e^t$ right?
You are correct that $\displaystyle A=1$ but then

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

$\displaystyle x(0)=e^{0}=1 \ne 0$ as is needed by you intial condition

5. Originally Posted by TheEmptySet
You are correct that $\displaystyle A=1$ but then

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

$\displaystyle x(0)=e^{0}=1 \ne 0$ as is needed by you intial condition
Good thinking. Can you comment on how the $\displaystyle x$ in $\displaystyle x+1$ turns into the function x(t)? It's making me uneasy.