MHT CET · Maths · Differential Equations
The particular solution of the differential equation \(\frac{d y}{d x}-e^x=y e^x\), when \(x=0\) and \(y=1\) is
- A \(\log \left(\frac{y+1}{2}\right)=\frac{e^x}{2}-\frac{1}{2}\)
- B \(\log \left(\frac{y+1}{2}\right)=e^x-1\)
- C \(\log (y-1)=e^x-1\)
- D \(\log 2(y-1)=e^x-1\)
Answer & Solution
Correct Answer
(B) \(\log \left(\frac{y+1}{2}\right)=e^x-1\)
Step-by-step Solution
Detailed explanation
\(\frac{d y}{d x}-e^x=y e^x \Rightarrow \frac{d y}{d x}=(y+1) e^x\) \( \Rightarrow \int \frac{d y}{y+1}=\int e^x d x \)
\( \Rightarrow \log (y+1)=e^x+c \)
\( \text { for } x=0, y=1 \Rightarrow c=\log 2-1 \)
\( \text { Hence, } \log (y+1)=e^x+\log 2-1 \)
\( \Rightarrow \log (y+1)-\log 2=e^x-1 \)
\( \Rightarrow \log \left(\frac{y+1}{2}\right)=e^x-1\)
\( \Rightarrow \log (y+1)=e^x+c \)
\( \text { for } x=0, y=1 \Rightarrow c=\log 2-1 \)
\( \text { Hence, } \log (y+1)=e^x+\log 2-1 \)
\( \Rightarrow \log (y+1)-\log 2=e^x-1 \)
\( \Rightarrow \log \left(\frac{y+1}{2}\right)=e^x-1\)
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