# Thread: Second Derivative with only "y" variable

1. ## Second Derivative with only "y" variable

How do you take the 2nd derivative for a function without an x on the right side of the equation?

Ie: y'= 0.05y with y(0)=1000 which would yield y'(0)=50, right?

How do I find y'' for this equation?

Thank you,
Jen

2. Originally Posted by Jenberl
How do you take the 2nd derivative for a function without an x on the right side of the equation?

Ie: y'= 0.05y with y(0)=1000 which would yield y'(0)=50, right?

How do I find y'' for this equation?

Thank you,
Jen

Hmm...this is what I think you should do.

I suggest you solve the differential equation $\frac{\,dy}{\,dx}=0.05y$ with the initial condition $y(0)=1000$

By separation of variables, this leads to $\frac{\,dy}{y}=0.05\,dx$

Integrate both sides, and we see that $\int\frac{\,dy}{y}=0.05\int\,dx\implies \ln\left|y\right|=0.05x+C\implies\left|y\right|=e^ {0.05x+C}$ $\implies \left|y\right|=e^Ce^{0.05x}\implies y=Ke^{0.05x};~K=\pm e^{C}$

Now applying the initial condition $y(0)=1000$, you get $1000=Ke^{0.05\cdot0}\implies K=1000$

Thus, your function is $y\!\left(x\right)=1000e^{0.05x}$

Now go ahead and find $\frac{\,d^2y}{\,dx^2}$

Does this make sense? Can you take it from here?

3. ## Second Derivative with only "y" variable- Taylor Series Needed

Originally Posted by Chris L T521
Hmm...this is what I think you should do.

I suggest you solve the differential equation $\frac{\,dy}{\,dx}=0.05y$ with the initial condition $y(0)=1000$

By separation of variables, this leads to $\frac{\,dy}{y}=0.05\,dx$

Integrate both sides, and we see that $\int\frac{\,dy}{y}=0.05\int\,dx\implies \ln\left|y\right|=0.05x+C\implies\left|y\right|=e^ {0.05x+C}$ $\implies \left|y\right|=e^Ce^{0.05x}\implies y=Ke^{0.05x};~K=\pm e^{C}$

Now applying the initial condition $y(0)=1000$, you get $1000=Ke^{0.05\cdot0}\implies K=1000$

Thus, your function is $y\!\left(x\right)=1000e^{0.05x}$

Now go ahead and find $\frac{\,d^2y}{\,dx^2}$

Does this make sense? Can you take it from here?

Well, yes...thank you. I wish I could do it that way, but I forgot to mention that the answer must be computed using the Taylor series method.
I'm still not sure how to create derivatives up to the 4th power to make this equation work with Taylor series. Any suggestions?