MHT CET · Maths · Differentiation
If \(\frac{\mathrm{d} y}{\mathrm{~d} x}=y+3, y+3\gt0\) and \(y(0)=2\), then \(y(\log 2)\) is equal to
- A 13
- B \(\quad-2\)
- C 7
- D 5
Answer & Solution
Correct Answer
(C) 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
\(\int \frac{\mathrm{d} y}{y+3}=\int \mathrm{d} x+\mathrm{c}\)
\(\Rightarrow \log (y+3)=x+\mathrm{c} ...(i)\)
\( y=2 \text { when } x=0 \)
\( \therefore \log (2+3)=0+\mathrm{c} \Rightarrow \mathrm{c}=\log 5 \)
\( \therefore \log (y+3)=x+\log 5...[From(i)] \)
\( \Rightarrow y+3=5 \mathrm{e}^x \)
\( \Rightarrow y=5 \mathrm{e}^x-3 \)
\( \therefore \quad y(\log 2)=5 \mathrm{e}^{\log 2}-3=10-3=7\)
...[From (i)]
& \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
\(\int \frac{\mathrm{d} y}{y+3}=\int \mathrm{d} x+\mathrm{c}\)
\(\Rightarrow \log (y+3)=x+\mathrm{c} ...(i)\)
\( y=2 \text { when } x=0 \)
\( \therefore \log (2+3)=0+\mathrm{c} \Rightarrow \mathrm{c}=\log 5 \)
\( \therefore \log (y+3)=x+\log 5...[From(i)] \)
\( \Rightarrow y+3=5 \mathrm{e}^x \)
\( \Rightarrow y=5 \mathrm{e}^x-3 \)
\( \therefore \quad y(\log 2)=5 \mathrm{e}^{\log 2}-3=10-3=7\)
...[From (i)]
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