MHT CET · Maths · Differential Equations
The differential equation \(\frac{d y}{d x}=\frac{\sqrt{1-y^2}}{y}\) determines a family of circles with
- A variable radii and a fixed centre at \((0,-1)\)
- B fixed radius of 1 unit and variable centres along the \(\mathrm{X}\)-axis
- C fixed radius of 1 unit and variable centres along the Y-axis
- D variable radii and a fixed centre at \((0,1)\)
Answer & Solution
Correct Answer
(B) fixed radius of 1 unit and variable centres along the \(\mathrm{X}\)-axis
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \frac{d y}{d x}=\frac{\sqrt{1-y^2}}{y} \\ & \Rightarrow \int \frac{y d y}{\sqrt{1-y^2}}=\int d x \\ & \Rightarrow-\sqrt{1-y^2}=x+c \\ & \Rightarrow 1-y^2=(x+c)^2 \\ & \Rightarrow(x+c)^2+y^2=1\end{aligned}\)
The above equation represents a circle having centre at \((-c, 0)\) which is variable and radius is equal to ' 1 ' which is fixed.
The above equation represents a circle having centre at \((-c, 0)\) which is variable and radius is equal to ' 1 ' which is fixed.
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