If #dy/dx = 6/(2x - 3)^2#, and #y(3) = 5#, what is the x-intercept of #y#?
1 Answer
Explanation:
We will have to solve a differential equation to find the initial function. This d.e. is separable.
#dy/dx = 6/(2x -3)^2#
#dy = 6/(2x - 3)^2dx#
Integrate both sides.
#intdy = int 6/(2x - 3)^2 dx#
The right-hand side can be integrated by a u-substitution. Let
#intdy =int 6/u^2 * (du)/2#
#intdy = int 3/u^2 du#
#intdy = 3u^-2#
#y = -3u^-1+ C#
#y = -3/u + C#
#y = -3/(2x - 3) + C#
Now solve for
#5 = -3/(2(3) - 3) + C#
#5 = -3/3 + C#
#5 = C - 1#
#C = 6#
Therefore, the initial function is
#0 = -3/(2x - 3) + 6#
#-6 = -3/(2x - 3)#
#-6(2x - 3) = -3#
#-12x + 18 = -3#
#-12x = -21#
#x = 21/12#
#x = 7/4#
Hopefully this helps!