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Example · Example 6

Q.Show that the differential equation (x2+y2) dx=2xy dy(x^2+y^2)\,dx = 2xy\,dy is homogeneous, and find its general solution.

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Write the equation as dydx=x2+y22xy\dfrac{dy}{dx} = \dfrac{x^2+y^2}{2xy}. Both x2+y2x^2+y^2 and 2xy2xy are homogeneous of degree 22 (each term has total degree 22 in x,yx,y), so the equation is homogeneous.

Substitute y=vxy=vx, so dydx=v+xdvdx\dfrac{dy}{dx}=v+x\dfrac{dv}{dx}:

v+xdvdx=x2+v2x22x⋅vx=1+v22v⟹xdvdx=1+v22v−v=1−v22v.v + x\frac{dv}{dx} = \frac{x^2+v^2x^2}{2x\cdot vx} = \frac{1+v^2}{2v} \quad\Longrightarrow\quad x\frac{dv}{dx} = \frac{1+v^2}{2v}-v = \frac{1-v^2}{2v}.

Separate:

2v1−v2 dv=dxx.\frac{2v}{1-v^2}\,dv = \frac{dx}{x}.

Integrate the left side using u=1−v2u=1-v^2, du=−2v dvdu=-2v\,dv: ∫2v1−v2 dv=−ln⁡∣1−v2∣\displaystyle\int\frac{2v}{1-v^2}\,dv = -\ln|1-v^2|. So

−ln⁡∣1−v2∣=ln⁡∣x∣+C1⟹(1−v2) x=C (renaming the constant).-\ln|1-v^2| = \ln|x| + C_1 \quad\Longrightarrow\quad (1-v^2)\,x = C \ \text{(renaming the constant)}. …

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