integrate dydx=(4sqrt(y)lnx)/x.
Check my answer? I got y=((2ln^2(x)+c)/2)^2. Is that right?
Take your answer, and take the derivative.
y=((2ln^2(x)+c)/2)^2
y'=2(sqrty)(4lnx/2)
Yep, looks right
I assume
dy/dx=(4sqrt(y)ln x)/x
y^-.5 dy = 4 (ln x /x)dx
y^.5 = 2 [ .5 ln^2 x ]
y^.5 = ln^2 x
y = ln^4 x + c
Yes we agree but I simplified
To check your answer, let's differentiate it and see if it matches the original differential equation.
Given:
dy/dx = (4√(y)ln(x))/x
Your proposed solution:
y = ((2ln^2(x) + c)/2)^2
Let's differentiate y with respect to x and see if we get the original equation:
dy/dx = d/dx(((2ln^2(x) + c)/2)^2)
Using the chain rule, we have:
dy/dx = 2((2ln^2(x) + c)/2)(d/dx(2ln^2(x) + c)/2)
Taking the derivative of ln^2(x) with respect to x, we get:
dy/dx = 2((2ln^2(x) + c)/2)(2ln(x) * (1/x))
Simplifying further, we have:
dy/dx = (2ln^2(x) + c)(ln(x)/x)
From this differentiation, we can see that the result does not match the original equation, which is dy/dx = (4√(y)ln(x))/x.
Therefore, your proposed solution for y, y = ((2ln^2(x) + c)/2)^2, is not correct.