solve the differential equation dy/dx=(3x*exp(4y))/(7+(2x^(2))^(2) when y = 0, x = 2

Move y terms to the left hand side and the keep the x terms on the right hand side: dy*(1/(exp(4y))) = dx*(3x/(7+2x^(2))^(2))
First, solve the right hand side by using a substition: u = 7 + 2x^(2), du/dx=4xFrom that the integral goes to (3/4)(u^(-2))du = -(3/4)(u^(-1)) + C = - ( 3/4)(1/(7+2x^(2))) + C
Second, solve the left hand side: dy(1/(exp(4y))) = dy*(exp(-4y)) = -(1/4y)exp(-4y)
exp(-40) = 1 = (3)(1/(7+2^(3))) + C, rearrange to find C = 1/5
So, exp(-4y) = 3/(7+2x^(2)) + 4/5,
y = -(1/4)ln((3/(7+2x^(2))) + 4/5) = ln((3/(7+2*x^(2))) + 4/5)^(-1/4)