Question

\(x^2\,dy+\dfrac12(xy+y^2)\,dx=0\); find the particular solution, given that \(y=1\) when \(x=1\).

Hint:

Homogeneous equation: substitute y=vx, separate variables, then apply y(1)=1.

Answer 1

Step 1: Rewriting as dy/dx:
\(x^2\,dy=-\dfrac12(xy+y^2)\,dx\Rightarrow \dfrac{dy}{dx}=-\dfrac{xy+y^2}{2x^2}\), which is homogeneous (degree 0 in x,y).

Step 2: Substituting y = vx:
Let \(y=vx\), so \(\dfrac{dy}{dx}=v+x\dfrac{dv}{dx}\). Then \(v+x\dfrac{dv}{dx}=-\dfrac{vx^2+v^2x^2}{2x^2}=-\dfrac{v+v^2}{2}\).

Step 3: Isolating x dv/dx:
\(x\dfrac{dv}{dx}=-\dfrac{v+v^2}{2}-v=-\dfrac{3v+v^2}{2}=-\dfrac{v(v+3)}{2}\).

Step 4: Separating and integrating:
\(\dfrac{dv}{v(v+3)}=-\dfrac{dx}{2x}\). Using partial fractions \(\dfrac{1}{v(v+3)}=\dfrac13\left(\dfrac1v-\dfrac1{v+3}\right)\): \(\dfrac13\ln\left|\dfrac{v}{v+3}\right|=-\dfrac12\ln|x|+C_1\), i.e. \(\dfrac{v}{v+3}=Kx^{-3/2}\).

Step 5: Applying the initial condition:
At \(x=1,y=1\): \(v=1\), so \(\dfrac{1}{4}=K(1)\Rightarrow K=\dfrac14\).

Step 6: Back-substituting v = y/x:
\(\dfrac{y/x}{y/x+3}=\dfrac14 x^{-3/2}\Rightarrow \dfrac{y}{y+3x}=\dfrac{1}{4}x^{-3/2}\Rightarrow 4y\,x^{3/2}=y+3x\Rightarrow y(4x^{3/2}-1)=3x\).

Final Answer:
\[ \boxed{y=\dfrac{3x}{4x^{3/2}-1}} \]