Question:medium

The initial value problem \[ (x-x^2)\frac{dy}{dx} = (2x-1)y, \qquad y(x_0)=y_0, \] has a unique solution if \((x_0,y_0)\) equals to

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For a first-order linear differential equation, \[ \boxed{ \frac{dy}{dx}+P(x)y=Q(x), } \] a unique solution exists wherever \(P(x)\) and \(Q(x)\) are continuous.
Updated On: Jul 14, 2026
  • \((0,0)\)
  • \((0,1)\)
  • \((1,1)\)
  • \((2,1)\)
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The Correct Option is D

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