How to solve z'(t)=1/2*z(t)

[mazharhussain]

Hi everyone,

I am a student of molecular biology. I had been working on my course of mathematical biology and I came across this ODE (z'(t)=1/2*z(t)). I don't know how to solve this. The answer to this problem is: z(t) = C*e^((1/2)*t)

I shall appreciate if someone give me step by step solution to this.

Best Regards
Mazhar

 

[Ishuda]

Hi everyone,

I am a student of molecular biology. I had been working on my course of mathematical biology and I came across this ODE (z'(t)=1/2*z(t)). I don't know how to solve this. The answer to this problem is: z(t) = C*e^((1/2)*t)

I shall appreciate if someone give me step by step solution to this.

Best Regards
Mazhar

Write z'(t)=1/2*z(t) as
\(\displaystyle \frac{dz}{dt} = \frac{1}{2} z\)
and multiply though by dt and divide by z to get
\(\displaystyle \frac{dz}{z} = \frac{1}{2} dt\)
Now integrate and simplify the answer. If you have problems show your work so we will know where to help.

 

[mazharhussain]

Write z'(t)=1/2*z(t) as
[FONT=MathJax_Math-italic]d[/FONT][FONT=MathJax_Math-italic]z[/FONT][FONT=MathJax_Math-italic]d[/FONT][FONT=MathJax_Math-italic]t[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Math-italic]z[/FONT]
and multiply though by dt and divide by z to get
[FONT=MathJax_Math-italic]d[/FONT][FONT=MathJax_Math-italic]z[/FONT][FONT=MathJax_Math-italic]z[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Math-italic]d[/FONT][FONT=MathJax_Math-italic]t[/FONT]
Now integrate and simplify the answer. If you have problems show your work so we will know where to help.


Thank you Ishuda,

I had come upto this point before. Let me solve it to point, from where I am needing explanation
Now integrating the expression (provided by you):

now, I can understand that when I will de-logarize the above expression then (1/2)t will be raised to the power of e. But I am not getting how log_eC multiplying factor to (1/2)t by giving

Pardon me for my non-technical language, for I am a Biologist. And thank you again for your help