How to solve the separable differential equation and to the initial condition: y(0)=1 ?

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1 Answer
May 6, 2016

Get the xs on one side and ys on the other, integrate, and simplify to get y=sqrt(5/2(sqrt(x^2+1))-3/2).

Explanation:

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).