Given that $y=1.5$ at $x=-2$, solve the differential equation $$\frac{dy}{dx}=\frac{\sqrt{4y+3}}{x^3}$$ giving your answer in the form $y=f(x)$. - Edexcel - A-Level Maths Pure - Question 7 - 2011 - Paper 5

We rearrange the equation to isolate variables:

$\int (4y + 3)^{-\frac{1}{2}} dy = \int \frac{1}{x^3} dx$

Which we can solve as:

1. The left side yields:

$\frac{1}{2}(4y + 3)^{\frac{1}{2}} = -\frac{1}{2x^2} + C$

2. At this point, we now substitute the initial conditions $(-2, 1.5)$ into the integrated equation to find ( C ).

Substituting:\n
$\frac{1}{2}(4(1.5) + 3) = -\frac{1}{2(-2)^2} + C$

This further simplifies to:

$\frac{1}{2}(6) = -\frac{1}{8} + C$

Hence,

$3 = -\frac{1}{8} + C$

Solving for ( C ) yields:

$C = 3 + \frac{1}{8} = \frac{25}{8}$

3. Substituting back, we have:

$\frac{1}{2}(4y + 3)^{\frac{1}{2}} = -\frac{1}{2x^2} + \frac{25}{8}$

4. Finally, rearranging the equation gives:

$y = \frac{1}{4} \left( \frac{2}{x} \right)^2 - \frac{3}{4}$.

So the solution can be expressed as:

$y = \frac{3-2x}{4x^2}$.