JEE Mains · Maths · STD 12 - 9. differential equations
If \(\frac{d y}{d x}=\frac{2^{x} y+2^{y} \cdot 2^{x}}{2^{x}+2^{x+y} \log _{e} 2}, y(0)=0\), then for \(y=1\) the value of \(x\) lies in the interval:
- A \((1,2)\)
- B \(\left(\frac{1}{2}, 1\right]\)
- C \((2,3)\)
- D \(\left(0, \frac{1}{2}\right]\)
Answer & Solution
Correct Answer
(A) \((1,2)\)
Step-by-step Solution
Detailed explanation
\(\frac{d y}{d x}=\frac{2^{x}\left(y+2^{y}\right)}{2^{x}\left(1+2^{y} \ell n 2\right)}\) \(\Rightarrow \int \frac{\left(1+2^{y}\right) \ell n 2}{\left(y+2^{y}\right)} d y=\int \,d x\) \(\Rightarrow \ell\) nly \(+2^{y} \mid=x+c\) \(x=0 ; y=0 \Rightarrow c=0\)…
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