MHT CET · Maths · Differential Equations
The general solution of the differential equation \(\frac{d y}{d x}=2^{y-x}\) is
- A \(2^x-2^y=c\)
- B \(\frac{1}{2^x}-\frac{1}{2^y}=c\)
- C \(\frac{1}{2^x}+\frac{1}{2^y}=c\)
- D \(2^x+2^y=c\)
Answer & Solution
Correct Answer
(B) \(\frac{1}{2^x}-\frac{1}{2^y}=c\)
Step-by-step Solution
Detailed explanation
\(\frac{d y}{d x}=2^{y-x}\)
\(\therefore \frac{d y}{d x}=\frac{2^y}{2^x} \Rightarrow \int \frac{d y}{2^y}=\int \frac{d x}{2^x}\)
\(\therefore \int 2^{-y} d y=\int 2^{-x} d x \Rightarrow \frac{2^{-y}}{-\ell n 2}=\frac{2^{-x}}{-\ln 2}+c_1 \Rightarrow\) \(\frac{2^{-y}}{\ln 2}=\frac{2^{-x}}{\ln 2}-c_1\)
\(\therefore \frac{1}{\ln 2}\left[2^{-x}-2^{-y}\right]=c_1\)
\(\therefore \frac{1}{2^x}-\frac{1}{2^y}=c, \text { where } c=\left(c_1\right)(\ln 2)\)
\(\therefore \frac{d y}{d x}=\frac{2^y}{2^x} \Rightarrow \int \frac{d y}{2^y}=\int \frac{d x}{2^x}\)
\(\therefore \int 2^{-y} d y=\int 2^{-x} d x \Rightarrow \frac{2^{-y}}{-\ell n 2}=\frac{2^{-x}}{-\ln 2}+c_1 \Rightarrow\) \(\frac{2^{-y}}{\ln 2}=\frac{2^{-x}}{\ln 2}-c_1\)
\(\therefore \frac{1}{\ln 2}\left[2^{-x}-2^{-y}\right]=c_1\)
\(\therefore \frac{1}{2^x}-\frac{1}{2^y}=c, \text { where } c=\left(c_1\right)(\ln 2)\)
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