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Question 44 of 50

Q.Solve (x² + y²) dx - 2xy dy = 0, given y = 0, when x = 1.

West Bengal WbchseWest Bengal HS (WBCHSE) Board 2024Subjective· 5mImportance★★★★★
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This is a homogeneous differential equation; substitute y=vxy=vx to separate the variables, then apply the initial condition.

Given (x2+y2) dx−2xy dy=0(x^2+y^2)\,dx - 2xy\,dy = 0, given y=0y=0 when x=1x=1.

Rewrite: dydx=x2+y22xy\dfrac{dy}{dx} = \dfrac{x^2+y^2}{2xy}. Both numerator and denominator are homogeneous of degree 22, so substitute y=vxy=vx, giving dydx=v+xdvdx\dfrac{dy}{dx} = v+x\dfrac{dv}{dx}:

v+xdvdx=x2+v2x22x⋅vx=1+v22vv+x\frac{dv}{dx} = \frac{x^2+v^2x^2}{2x\cdot vx} = \frac{1+v^2}{2v}

xdvdx=1+v22v−v=1+v2−2v22v=1−v22vx\frac{dv}{dx} = \frac{1+v^2}{2v}-v = \frac{1+v^2-2v^2}{2v} = \frac{1-v^2}{2v}

Separate variables:

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

Integrate. On the left, since ddv(1−v2)=−2v\dfrac{d}{dv}(1-v^2)=-2v, we get ∫2v1−v2dv=−ln⁡∣1−v2∣\displaystyle\int\frac{2v}{1-v^2}dv = -\ln|1-v^2|:

−ln⁡∣1−v2∣=ln⁡∣x∣+C1-\ln|1-v^2| = \ln|x| + C_1

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