MHT CET · Maths · Differential Equations
If \(\frac{\mathrm{d} y}{\mathrm{~d} x}=y+3\) and \(y(0)=2\), then \(y(\log 2)=\)
- A 5
- B 7
- C 13
- D -2
Answer & Solution
Correct Answer
(B) 7
Step-by-step Solution
Detailed explanation
\(
\begin{aligned}
& \frac{\mathrm{d} y}{\mathrm{~d} x}=y+3 \\
& \Rightarrow \frac{\mathrm{d} y}{y+3}=\mathrm{d} x
\end{aligned}
\)
Integrating on both sides, we get
\( \int \frac{\mathrm{d} y}{y+3}=\int \mathrm{d} x+\mathrm{c} \)
\( \Rightarrow \log (y+3)=x+\mathrm{c} \)
\( y=2 \text { when } x=0 \)
\( \therefore \log (2+3)=0+\mathrm{c} \Rightarrow \mathrm{c}=\log 5 \)
\( \therefore \log (y+3)=x+\log 5 \)
\( \Rightarrow y+3=5 \mathrm{e}^x \)
\( \Rightarrow y=5 \mathrm{e}^x-3 \)
\( \therefore y(\log 2)=5 \mathrm{e}^{\log 2}-3=10-3 \)
\( =7\)
\begin{aligned}
& \frac{\mathrm{d} y}{\mathrm{~d} x}=y+3 \\
& \Rightarrow \frac{\mathrm{d} y}{y+3}=\mathrm{d} x
\end{aligned}
\)
Integrating on both sides, we get
\( \int \frac{\mathrm{d} y}{y+3}=\int \mathrm{d} x+\mathrm{c} \)
\( \Rightarrow \log (y+3)=x+\mathrm{c} \)
\( y=2 \text { when } x=0 \)
\( \therefore \log (2+3)=0+\mathrm{c} \Rightarrow \mathrm{c}=\log 5 \)
\( \therefore \log (y+3)=x+\log 5 \)
\( \Rightarrow y+3=5 \mathrm{e}^x \)
\( \Rightarrow y=5 \mathrm{e}^x-3 \)
\( \therefore y(\log 2)=5 \mathrm{e}^{\log 2}-3=10-3 \)
\( =7\)
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