Expert: Alon Mandes - 6/30/2010

Question
given ∫ (f(x))^n f'(x)dx = 1/n+1(f(x))^n+1 +c (n does not = -1)?

1. Using the above equation show that
∫ [(cos x) / √(4 - sin x) ]dx = -2√(4 - sin x) + c
where c is an arbitrary constant

2. Hence find, in implicit form, the general solution of the differential equation
dy/dx = (5y^4/5 cos x)/√(4 - sin x) (y>0)

3. Find the corresponding particular solution (in implicit form) that satisfies the initial condition y=32 when x=0

4. Find the explicit form of this particular solution.

any help welcome, thanks

xxx

Answer
1.
Let's consider the integral : ∫ [ (cosx) / √(4-sinx) ]dx . We set f(x)=4-sinx , we get :
f'(x)=-cosx . Thus , the integral will become :
∫ [ -f'(x)/√f(x) ]dx = ∫ -[f(x)^(-1/2) ]f'(x) dx . According to the rule above, the solution
is :
{-1/(1+[-1/2]) } * f(x)^[1-(1/2)] = -2f(x)^(1/2)=-2√f(x)=-2√(4-sinx)+c .


2.
We have the DE : dy/dx=[5y^(4/5)]*cosx/√(4-sinx) . Let's separate x & y , we get :
dy/5y^(4/5) = [cosx/√(4-sinx)] dx
(1/5)y^(-4/5) dy = [cosx/√(4-sinx)] dx
Now let's integrate both sides with respect to x & y accordingly , we get :
∫ (1/5)y^(-4/5) dy = ∫ [cosx/√(4-sinx)] dx
The right side we solved it in (1) . All we need to do is to solve the left side :
(1/5)*[1/1-(4/5)]y^(1-[4/5]) + c = -2√(4-sinx)+c
The solution :
y^(1/5) + 2√(4-sinx) = C

Now, let's apply the initial conditions : x=0 --> y=32 :
32^(1/5)+2√4 = C --> 2+4=C --> C=6 .

Alon.