Answer:
(1+y)^(3/2)-3(1+y)^(1/2)=(1+x)^(3/2)-3(1+x)^(1/2)+c
Step-by-step explanation:
This is separable.
Multiply x dx on both sides and divide both sides by (1+y)^(1/2) to obtain:
y/(1+y)^(1/2)=(1+x)^(-1/2) x dx
Using law of exponents to get rid of negative exponent by finding reciprocal:
y/(1+y)^(1/2)=x/(1+x)^(1/2) dx
Let u=(1+y)^(1/2).
Then u^2=1+y.
Differentiate: 2 u du=dy
[Let v=(1+x)^(1/2).]
Plug in our substitution:
(u^2-1) 2u du/u=(v^2-1) 2v dv/v
Simplify:
(u^2-1) 2 du=(v^2-1) 2 dv
Integrate both sides:
(u^3/3-u) 2+L=(v^3/3-v) 2+K
Divide both sides by 2:
u^3/3-u +L/2=v^3/3-v+K/2
K/2-L/2 is just a constant value C:
u^3/3-u=v^3/3-v+C
Use substitution again to put back in terms of x and y:
((1+y)^(1/2))^3/3-(1+y)^(1/2)=((1+x)^(1/2))^3/3-(1+x)^(1/2)+C
Simplify the exponent part a bit:
(1+y)^(3/2)/3-(1+y)^(1/2)=(1+x)^(3/2)/3-(1+x)^(1/2)+C
I don't like the division by 3 so multiply both sides by 3:
(1+y)^(3/2)-3(1+y)^(1/2)=(1+x)^(3/2)-3(1+x)^(1/2)+3C
3C is still just an unknown constant, c.
(1+y)^(3/2)-3(1+y)^(1/2)=(1+x)^(3/2)-3(1+x)^(1/2)+c
