First order differential equation

[Mondo]

Hello,

I try to solve this equation [math]\frac{dX}{ds} = -\frac{X}{R}[/math] I separate variables and calculate two separate integrals like this [math]\frac{dX}{X} = -\frac{ds}{R} \rightarrow ln(X) + C_1 = -\frac{s}{R} + C_2 \rightarrow X = e^{-\frac{s}{R}} + e^{C_2} - e^{C_1}[/math]
But this is not the right answer. What do I do wrong here?

 

[skeeter]

[imath]\ln(X) = -\dfrac{s}{R} - C_1 + C_2[/imath]

[imath]X = e^{-\frac{s}{R} - C_1 + C_2} \ne e^{-\frac{s}{R}} - e^{C_1} + e^{C_2}[/imath]

 

[blamocur]

You are assuming that [imath]e^{a+b} = e^a + e^b[/imath], which is not true.

 

[Mondo]

[imath]\ln(X) = -\dfrac{s}{R} - C_1 + C_2[/imath]

[imath]X = e^{-\frac{s}{R} - C_1 + C_2} \ne e^{-\frac{s}{R}} - e^{C_1} + e^{C_2}[/imath]

Thank you, I see my mistake now. But the answer is yet different - it is [math]x = Re^{-s/R}[/math]
Why is that?

 

[skeeter]

[imath]X = e^{-\frac{s}{R} + C_2 - C_1} = e^{C_2-C_1} \cdot e^{-\frac{s}{R}} = C_3 \cdot e^{-\frac{s}{R}}[/imath]

were you given an initial condition to determine [imath]C_3 = R[/imath] ?

 

[Mondo]

Yes @skeeter, I just used them and got the right answer. Thank you!