1.

Solve the following differential equation: \((\frac{e^z-1}{e^z})dx-(\frac{e^z+1}{e^z})dy=0\)(ez - 1/(ez)) * d * x - (ez + 1/(ez)) * d * y = 0

Answer»

\((\frac{e^2-1}{e^2})dx-(\frac{e^z+1}{e^2})dy=0\)

(1 - e-z)dx - (1 + e-z)dy = 0

∫(1 - e-z)dx - ∫(1 + e-z)dy = 0

Let z = x + iy

⇒ ∫(1 - e-(x + iy))dx - ∫(1 + e-(x + iy))dy = c

⇒ x + e-(x+iy) - y + \(\frac1{i}\)e-(x + iy) = c

⇒ x - y + e-(x + iy)(1 + \(\frac1i\times\frac{i}i\)) = c

⇒ x - y + e-z(1 - i) = c (\(\because\) i= -1)

Hence, solution of given differential equation is x - y + e-z(1 - i) = c



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