MHT CET · Maths · Differential Equations
The solution of the differential equation \(\mathrm{e}^{\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)}=x+1 ; y(0)=5\), \(x \in(-1, \infty)\)
- A \((x-1) \log (x+1)-x-5=y\)
- B \((x+1) \log (x+1)+x+5=y\)
- C \((x-1) \log (x+1)+x-5=y\)
- D \((x+1) \log (x+1)-x+5=y\)
Answer & Solution
Correct Answer
(D) \((x+1) \log (x+1)-x+5=y\)
Step-by-step Solution
Detailed explanation
\(\mathrm{e}^{\frac{\mathrm{d} y}{\mathrm{~d} x}}=x+1\)
\(\Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=\log _{\mathrm{e}}(x+1)\)
\(\Rightarrow \int \mathrm{d} y=\int \log _{\mathrm{e}}(x+1) \mathrm{d} x\)
\(\Rightarrow y=(x+1) \log _{\mathrm{e}}(x+1)-(x+1)+C \text { [integrating}\) \(\text{by parts] }\)
\(\text { putting } x=0 \text { and } y=5 \text { we get } C=6\)
\(\Rightarrow y=(x+1) \log _{\mathrm{e}}(x+1)-x+5\)
\(\Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=\log _{\mathrm{e}}(x+1)\)
\(\Rightarrow \int \mathrm{d} y=\int \log _{\mathrm{e}}(x+1) \mathrm{d} x\)
\(\Rightarrow y=(x+1) \log _{\mathrm{e}}(x+1)-(x+1)+C \text { [integrating}\) \(\text{by parts] }\)
\(\text { putting } x=0 \text { and } y=5 \text { we get } C=6\)
\(\Rightarrow y=(x+1) \log _{\mathrm{e}}(x+1)-x+5\)
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