setting up an integral

Question

set up an integral for solving dy/dx=1/(x^2-16) when y(0)=0

 

y(x)= ...... + (integral for ..... to ......) .........

 

 

Also Evaluate your answer to the previous part to find y(x)

 

y(x)=?

 

Fill the blanks.

 

This question has been annoying me for ages.

Answer 1

∫₀^xdy/dt=∫₀^x[1/(t²-16)]dt

y(x)-y(0)=∫₀^x[1/(t²-16)]dt

 

y(x)=y(0)+∫₀^x[1/(t²-16)]dt=(1/8)∫₀^x[1/(t-4)-1/(t+4)]dt

=(1/8)ln|(t-4)/(t+4)| |₀^x=(1/8)ln|(x-4)/(x+4)|+ln1

Since y(0)=0 and  ln1=0

 

y(x)=(1/8)ln|(x-4)/(x+4)|