Question:
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
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
Show Hint
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
- \((0,0)\)
- \((0,1)\)
- \((1,1)\)
- \((2,1)\)
Show Solution
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The Correct Option is D
Solution and Explanation
Step 1: Write the equation in standard form.
Dividing by \(x-x^2=x(1-x)\),
\[
\frac{dy}{dx}
=
\frac{2x-1}{x-x^2}\,y.
\]
The coefficient is continuous only when
\[
x\neq0,\;1.
\]
Step 2: Apply the existence and uniqueness theorem. A unique solution exists if the coefficient is continuous at the initial point. Among the given options, \[ x_0=2 \] is the only value satisfying \[ x_0\neq0,\;1. \] Hence, \[ \boxed{(2,1)} \] is the correct initial point. Therefore, \[ \boxed{(D)} \] is the correct answer.
Step 2: Apply the existence and uniqueness theorem. A unique solution exists if the coefficient is continuous at the initial point. Among the given options, \[ x_0=2 \] is the only value satisfying \[ x_0\neq0,\;1. \] Hence, \[ \boxed{(2,1)} \] is the correct initial point. Therefore, \[ \boxed{(D)} \] is the correct answer.
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