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Find the solution of `(dy)/(dx)=2^(y-x)` |
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Answer» `"Given that",=(dy)/(dx)=2^(y-x)` `Rightarrow (dy)/(dx)=(2^(y))/(2^(x))[therefore 2^(m-n)=(a^(m))/(a^(n))]` `Rightarrow (dy)/(2^(y))=(dx)/(2^(x))` On integrating both sides, we get `int2^(-y)dy=int2^(-x)dx` `Rightarrow (-2^(-y))/(log2)=(-2^(-x))/(log2)+C` `Rightarrow -2^(-y)+2^(-x)=+Clog2` `Rightarrow 2^(-x)+2^(-y)=-Clog2` `Rightarrow 2^(-x)+2^(-y)=K` |
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