JEE Mains · Maths · STD 12 - 9. differential equations
Let \(y=y_{1}(x)\) and \(y=y_{2}(x)\) be two distinct solutions of the differential equation \(\frac{d y}{d x}=x+y\), with \(y _{1}(0)=0\) and \(y _{2}(0)=1\) respectively. Then, the number of points of intersection of \(y=y_{1}(x)\) and \(y=y_{2}(x)\) is.
- A \(0\)
- B \(1\)
- C \(2\)
- D \(3\)
Answer & Solution
Correct Answer
(A) \(0\)
Step-by-step Solution
Detailed explanation
\(\frac{d y}{d x}=x+y \Rightarrow \frac{d y}{d x}-y=x\) \(\text { 1f }= e ^{-x}\) \(\therefore\) solution is \(y e^{-x}=\int x e^{-x} d x\) \(y e^{-x}=-x e^{-x}-e^{-x}+c\) \(y=-x-1+c e^{x}\) \(y_{1}(0)=0 \Rightarrow c=1\) \(\therefore y_{1}=-x-1+e^{x}\)…
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