MHT CET · Maths · Differential Equations
The particular solution of \(\log \left(\frac{d y}{d x}\right)=3 x+4 y\) at \(x=y=0\) is
- A \(3 e^{-4 y}-4 e^{3 x}=7\)
- B \(3 e^{-4 y}+4 e^{3 x}=7\)
- C \(4 e^{-4 y}-3 e^{3 x}=7\)
- D \(4 e^{-4 y}+3 e^{3 x}=7\)
Answer & Solution
Correct Answer
(B) \(3 e^{-4 y}+4 e^{3 x}=7\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \log \left(\frac{d y}{d x}\right)=3 x+4 y \text { and } x=y=0 \\ & \Rightarrow \frac{d y}{d x}=e^{3 x+4 y} \\ & \Rightarrow \int e^{-4 y} d y=\int e^{3 x} d x \\ & \Rightarrow \frac{e^{-4 y}}{-4}=\frac{e^{3 x}}{3}+C\end{aligned}\)
Putting \(x=0\) and \(y=0\) we get \(C=\frac{-7}{12}\)
\(\begin{aligned} & \frac{e^{-4 y}}{-4}=\frac{e^{3 x}}{3}-\frac{7}{12} \\ & \Rightarrow 3 e^{-4 y}=-4 e^{3 x}+7 \\ & \Rightarrow 3 e^{-4 y}+4 e^{3 x}=7\end{aligned}\)
Putting \(x=0\) and \(y=0\) we get \(C=\frac{-7}{12}\)
\(\begin{aligned} & \frac{e^{-4 y}}{-4}=\frac{e^{3 x}}{3}-\frac{7}{12} \\ & \Rightarrow 3 e^{-4 y}=-4 e^{3 x}+7 \\ & \Rightarrow 3 e^{-4 y}+4 e^{3 x}=7\end{aligned}\)
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