AP EAMCET · Maths · Differential Equations
The equation of a curve passing through the point \((0,1)\), given that the slope of the tangent to the curve at any point \((x, y)\) is equal to the sum of the \(x\)-coordinate and the product of \(x\) and \(y\) coordinates at that point, is
- A \(y=1-2 e^{\left(\frac{x^2}{2}\right)}\)
- B \(y=-1+2 e^{\left(\frac{x^2}{2}\right)}\)
- C \(y=-1-2 e^{\left(\frac{x^2}{2}\right)}\)
- D \(y=1+2 e^{\left(\frac{x^2}{2}\right)}\)
Answer & Solution
Correct Answer
(B) \(y=-1+2 e^{\left(\frac{x^2}{2}\right)}\)
Step-by-step Solution
Detailed explanation
Given, \(\frac{d y}{d x}=x+x y\) \(\frac{d y}{d x}=x(1+y) \Rightarrow \frac{d y}{1+y}=x d x\) On integrating, we get \(\log (y+1)=\frac{x^2}{2}+C\) \(\Rightarrow \quad y+1=A e^{x^2 / 2} \Rightarrow y=A e^{x^2 / 2}-1\) Since it passes through \((0,1)\).…
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