Answer:
[tex]y=\frac{(-\frac{4}{x}+1)^2-3 }{4}[/tex]
Step-by-step explanation:
We are given the following information. y have the point [tex](-2,\frac{3}{2} )[/tex] and [tex]\frac{dy}{dx} =\frac{\sqrt{4y+3} }{x^2}[/tex]
First, we need to separate the variables to their respective sides
[tex]\frac{1}{\sqrt{4y+3} } dy=\frac{1}{x^2} dx[/tex]
Now, we need to integrate each side
[tex]\int \frac{1}{\sqrt{4y+3} } dy=\int\frac{1}{x^2} dx[/tex]
But first, let us rewrite these functions
[tex]\int (4y+3)^{-\frac{1}{2} } dy=\int x^{-2} dx[/tex]
Before we can integrate, we need to have the hook for the first function. When we integrate [tex](4y+3)^{-\frac{1}{2} }[/tex], we must have a lone 4 within the integral as well.
[tex]\frac{1}{4} \int4 (4y+3)^{-\frac{1}{2} } dy=\int x^{-2} dx[/tex]
Now we can integrate each side to get
[tex]\frac{1}{4} \sqrt{4y+3} =-\frac{1}{x} + c[/tex]
Now is the best time to use the given point in order to find the value of c.
[tex]\frac{1}{4} \sqrt{4(\frac{3}{2}) +3} =-\frac{1}{-2} + c\\\\\frac{1}{4}\sqrt{6+3} =\frac{1}{2} +c \\\\\frac{3}{4}=\frac{1}{2} +c\\ \\c=\frac{1}{4}[/tex]
Now we can plug in our value for c and then solve for y
[tex]\frac{1}{4} \sqrt{4y+3} =-\frac{1}{x} + \frac{1}{4} \\\\\sqrt{4y+3}=-\frac{4}{x} +1\\ \\4y+3=(-\frac{4}{x} +1)^2\\\\4y=(-\frac{4}{x} +1)^2-3\\\\y=\frac{(-\frac{4}{x} +1)^2-3}{4}[/tex]
