How can I solve this differential equation with given initial conditions?

[amcavoy]

I have the following to solve:

$$\frac{dx}{dt}=-\alpha xy;\quad y=y_0e^{-\beta t};\quad x(0)=x_0$$

I separate variables and come up with:

$$\frac{dx}{x}=-\alpha y_0e^{-\beta t}dt$$

$$\ln{x}=-\alpha y_0\int e^{-\beta t}dt=\frac{\alpha y_0}{\beta}e^{-\beta t}+C$$

...so for a final answer I come up with:

$$x=x_0\exp{\left(\frac{\alpha y_0}{\beta}e^{-\beta t}\right)}$$

..however the book says that the answer is:

$$x=x_0\exp{\left(\frac{-\alpha y_0\left(1-e^{-\beta t}\right)}{\beta}\right)}$$

I cannot find where I went wrong, any ideas?

Thanks a lot.

 

[Tide]

You didn't evaluate the time integral at both limits.

 

[hotvette]

Good catch. I stared at it for a few minutes and couldn't figure it out.

 

[amcavoy]

You didn't evaluate the time integral at both limits.

...both limits?

 

[saltydog]

You have:

$$x(t)=K\text{Exp}\left[\frac{\alpha y_0}{\beta}e^{-\beta t}\right]$$

with:

$$x(0)=x_0$$

Now, carefully substitute that initial value into the equation to solve for K.

 

[amcavoy]

You have:

$$x(t)=K\text{Exp}\left[\frac{\alpha y_0}{\beta}e^{-\beta t}\right]$$

with:

$$x(0)=x_0$$

Now, carefully substitute that initial value into the equation to solve for K.

Ahh, I must have put the x₀ term there prematurely. Thanks for the help everyone, I have it now.