We will begin to solve this first-order separable differential equation by separating it (no surprise there).
If we add #8ysqrt(x^2+1)dy/dx# to both sides we get:
#10x=8ysqrt(x^2+1)dy/dx#
Now divide by #8sqrt(x^2+1)# to get:
#(10x)/(8sqrt(x^2+1))=ydy/dx#
Multiply by #dx# to finally end up with:
#(10x)/(8sqrt(x^2+1))dx=ydy#
Yay! We've separated the equation: we have #x# on one side and #y# on the other. The only thing that's left is to integrate both sides:
#int(10x)/(8sqrt(x^2+1))dx=intydy#
Let's start with the more complicated one of these:
#int(10x)/(8sqrt(x^2+1))dx#
First, take out a #10/8# and simplify:
#5/4intx/(sqrt(x^2+1))dx#
Now we can apply the substitution #u=x^2+1->(du)/dx=2x->du=2xdx#
In order to apply the substitution, we need to multiply the inside of the integral by #2#, so we end up with #2x#. If we do that, we need to multiply the outside of the integral by #1/2#:
#5/8int(2x)/(sqrt(x^2+1))dx#
We can go ahead and substitute now:
#5/8int(du)/sqrt(u)#
Noticing that this is equivalent to #5/8intu^(-1/2)du# and using the reverse power rule, we end up with a solution of:
#2*5/8(u^(1/2))+C=5/4sqrt(u)+C#
Since #u=x^2+1#, we back-substitute to get our final answer:
#5/4sqrt(x^2+1)+C#
As for the other integral, #intydy#, well, that's just #y^2/2+C#.
Alright, we've solved our integrals so we now have:
#5/4sqrt(x^2+1)+C=y^2/2+C#
Doing a little algebra to solve for #y# yields:
#y=sqrt(5/2(sqrt(x^2+1))+C)#
Note: Remember that manipulating the integration constant #C# makes no difference. Whatever we do to it, it's still a constant.
Now we apply the initial condition #y(0)=1# to solve for #C#:
#y=sqrt(5/2(sqrt(x^2+1))+C)#
#1=sqrt(5/2(sqrt((0)^2+1))+C)#
#1=5/2+C#
#C=-3/2#
Therefore our solution is #y=sqrt(5/2(sqrt(x^2+1))-3/2)#.
