Question

Y/x dy/dx=(1+x^2)^-1/2(1+y^2)^1/2

Answer 1

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