MHT CET · Maths · Differential Equations
The solution of the differential equation \(\log \left(\frac{\mathrm{dy}}{\mathrm{d} x}\right)=9 x-6 \mathrm{y}+6\) is
(given that \(\mathrm{y}=1\) when \(x=0\) )
- A \(3 e^{6 y}=2 e^{9 x-6}+e^{6}\)
- B \(3 e^{6 y}=2 e^{9 x+6}+e^{6}\)
- C \(3 e^{6 y}=2 e^{9 x+6}-e^{6}\)
- D \(3 e^{6 y}=2 e^{9 x-6}-e^{6}\)
Answer & Solution
Correct Answer
(B) \(3 e^{6 y}=2 e^{9 x+6}+e^{6}\)
Step-by-step Solution
Detailed explanation
\(\log \left(\frac{d y}{d x}\right)=9 x-6 y+6 \Rightarrow \frac{d y}{d x}=e^{9 x-6 y+6}\)
\(\therefore \frac{d y}{d x}=e^{9 x+6} \cdot e^{-6 y} \Rightarrow \frac{d y}{e^{-6 y}}=e^{9 x} \cdot e^{6} d x\)
\(\therefore \int e^{6 y} d y=e^{6} \int e^{9 x} d x\)
\(\frac{e^{6 y}}{6}=\frac{e^{6} e^{9 x}}{9}+C\)
We have \(x=0, y=1\)
\(\therefore \frac{e^{6}}{6}=\frac{e^{6}}{9}+C \Rightarrow C=\frac{e^{6}}{18}\)
\(\therefore \frac{e^{6 y}}{6}=\frac{e^{6} e^{9 x}}{9}+\frac{e^{6}}{18} \Rightarrow 3 e^{6 y}=2 e^{6+9 x}+e^{6}\)
\(\therefore \frac{d y}{d x}=e^{9 x+6} \cdot e^{-6 y} \Rightarrow \frac{d y}{e^{-6 y}}=e^{9 x} \cdot e^{6} d x\)
\(\therefore \int e^{6 y} d y=e^{6} \int e^{9 x} d x\)
\(\frac{e^{6 y}}{6}=\frac{e^{6} e^{9 x}}{9}+C\)
We have \(x=0, y=1\)
\(\therefore \frac{e^{6}}{6}=\frac{e^{6}}{9}+C \Rightarrow C=\frac{e^{6}}{18}\)
\(\therefore \frac{e^{6 y}}{6}=\frac{e^{6} e^{9 x}}{9}+\frac{e^{6}}{18} \Rightarrow 3 e^{6 y}=2 e^{6+9 x}+e^{6}\)
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