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Question 39 of 40

Q.Solve the following differential equation:
(x2−y2)dx+2xy dy=0(x^2 - y^2)dx + 2xy\,dy = 0

Maharashtra MsbshseMaharashtra HSC (MSBSHSE) Board 2026Subjective· 4mImportance★★★★★
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The equation (x2−y2) dx+2xy dy=0(x^2 - y^2)\,dx + 2xy\,dy = 0 is homogeneous. Putting y=vxy = vx separates the variables; integrating gives x(1+v2)=Cx(1 + v^2) = C, and back-substituting v=y/xv = y/x yields x2+y2=Cxx^2 + y^2 = Cx.

Rewrite in the form dydx\frac{dy}{dx}:

(x2−y2) dx+2xy dy=0  ⇒  dydx=−(x2−y2)2xy=y2−x22xy(x^2 - y^2)\,dx + 2xy\,dy = 0 \;\Rightarrow\; \frac{dy}{dx} = \frac{-(x^2 - y^2)}{2xy} = \frac{y^2 - x^2}{2xy}

Since the right side is a function of y/xy/x (homogeneous of degree 00), substitute

y=vx,dydx=v+xdvdxy = vx, \qquad \frac{dy}{dx} = v + x\frac{dv}{dx}

Then

v+xdvdx=(vx)2−x22x(vx)=x2(v2−1)2vx2=v2−12vv + x\frac{dv}{dx} = \frac{(vx)^2 - x^2}{2x(vx)} = \frac{x^2(v^2 - 1)}{2vx^2} = \frac{v^2 - 1}{2v}

Isolate the derivative term:

xdvdx=v2−12v−v=v2−1−2v22v=−(1+v2)2vx\frac{dv}{dx} = \frac{v^2 - 1}{2v} - v = \frac{v^2 - 1 - 2v^2}{2v} = \frac{-(1 + v^2)}{2v}

Separate the variables:

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

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