InterviewSolution
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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 |
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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 (\(\because\) i2 = -1) Hence, solution of given differential equation is x - y + e-z(1 - i) = c |
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