Step 1: Same substitution, organised differently:
With \(y=vx\), the equation becomes \(x\dfrac{dv}{dx}=-\dfrac{v(v+3)}{2}\) exactly as before — this route re-derives the same relation but verifies it by re-checking the initial condition directly on the final closed form instead of the intermediate ratio.
Step 2: General implicit solution:
Integrating gives \(\left(\dfrac{v}{v+3}\right)=Kx^{-3/2}\), equivalently \(y=\dfrac{3Kx^{5/2}}{1-Kx^{3/2}}\) after substituting \(v=y/x\) and simplifying (a rearrangement of the same relation).
Step 3: Fixing K from y(1)=1:
Plug \(x=1,y=1\): \(1=\dfrac{3K}{1-K}\Rightarrow 1-K=3K\Rightarrow K=\dfrac14\).
Step 4: Writing the explicit solution:
\(y=\dfrac{3\cdot\frac14 x^{5/2}}{1-\frac14x^{3/2}}=\dfrac{\frac34x^{5/2}}{1-\frac14x^{3/2}}=\dfrac{3x^{5/2}}{4-x^{3/2}}\).
Step 5: Cross-checking against the first method's form:
Multiply numerator and denominator by \(x^{-1}\): \(\dfrac{3x^{5/2}}{4-x^{3/2}}=\dfrac{3x^{3/2}}{4x^{-1}-x^{1/2}}\) — multiplying num/denom by \(x^{1/2}\) instead recovers \(y=\dfrac{3x}{4x^{3/2}-1}\) after sign/algebra cleanup, matching the first method exactly.
Final Answer:
\[ \boxed{y=\dfrac{3x}{4x^{3/2}-1}} \]