KCET · Maths · Differential Equations
The solution of \(e^{d y / d x}=x+1, y(0)=3\) is
- A \(y-2=x \log x-x\)
- B \(y-x-3=x \log x\)
- C \(y-x-3=(x+1) \log (x+1)\)
- D \(y+x-3=(x+1) \log (x+1)\)
Answer & Solution
Correct Answer
(D) \(y+x-3=(x+1) \log (x+1)\)
Step-by-step Solution
Detailed explanation
\(\because e^{\frac{d y}{d x}}=x+1\)
\(\Rightarrow \quad \frac{d y}{d x}=\log (x+1)\)
\(\Rightarrow \quad \int d y=\int \log (x+1) d x\)
\(\Rightarrow \quad y=x \log (x+1)-\int \frac{x}{x+1} d x\)
\(\Rightarrow \quad y=x \log (x+1)-\int 1 \cdot d x+\int \frac{1}{1+x} d x\)
\(\Rightarrow \quad y=x \log (x+1)-x+\log (x+1)+C\)
\(\Rightarrow \quad y=(x+1) \log (x+1)-x+C\)
Now, \(y(0)=3\)
\(3=0+C \Rightarrow C=3\)
Hence, \(y=(x+1) \log (x+1)-x+3\)
\(y+x-3=(x+1) \log (x+1)\)
\(\Rightarrow \quad \frac{d y}{d x}=\log (x+1)\)
\(\Rightarrow \quad \int d y=\int \log (x+1) d x\)
\(\Rightarrow \quad y=x \log (x+1)-\int \frac{x}{x+1} d x\)
\(\Rightarrow \quad y=x \log (x+1)-\int 1 \cdot d x+\int \frac{1}{1+x} d x\)
\(\Rightarrow \quad y=x \log (x+1)-x+\log (x+1)+C\)
\(\Rightarrow \quad y=(x+1) \log (x+1)-x+C\)
Now, \(y(0)=3\)
\(3=0+C \Rightarrow C=3\)
Hence, \(y=(x+1) \log (x+1)-x+3\)
\(y+x-3=(x+1) \log (x+1)\)
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