Differential Equation (1st degree)?

Question

Given dy/dθ = y/(e^2θ+1), use the substitution x = e^θ to show that:
∫dy/y = ∫dx / (x(x^2+1))

How can I go about solving this? Any help would be appreciated, thanks.

Answer 1

Easy. You have θ = lnx, so dθ = dx/x. The substitution gives you...

dy/(dx/x) = y/(x²+1)
→ dy/y = dx/[x(x²+1)].

Answer 2

dy/y= d@/(e^2@ +1)
e^@=x so e^@d@= dx and d@= dx/x
so
Int dy/y = Int dx/(x^2+1)*x
The the left hand int is lnIyI+C
To calculate the right side put 1/x(x^2+1) = a/x +(bx+c)/(x^2+1)
so taking common denominator
1= a(x^2+1) +x(bx+c)
As both sides are identical give x arbitrary values
x=0 gives a=1
x=1 gives 1= 2+b+c
x=-1 gives 1= 2+b-c
sum 2= 4+2b so b=-1 and c= 0

so your integral is lnIxI -1/2 lnI x^2+1I
Take it from here on

Answer 3

x = e^ theta dx = e^theta dtheta

therefore dx/x = d (theta)

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

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

therefore integral dy/y = integral dx /x(x^2+1)
iwe can integrate it using partial fractions.
thankyou

Answer 4

(ill use T for theta)
first, separate the variables:
dy/y=dT/(e^2T+1)
Then you have dx/dT= e^T = x, so dT=dx/x
Now substitute and u get:
dy/y= dx/(x(x^2+1))
And then integrate both sides