InterviewSolution
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Solve the differential equation:x3\(\frac{d^3y}{dx^3}+3x^2\frac{d^2y}{dx^2}+x\frac{dy}{dx}+y=\) x + log x |
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Answer» x3\(\frac{d^3y}{dx^3}+3x^2\frac{d^2y}{dx^2}+x\frac{dy}{dx}+y=\) x + log x---(1) Let x = ez ⇒ log x = z Then x \(\frac{dy}{dx}\) converts into \(\frac{dy}{dx}\) = Dy where D = d/dz x2\(\frac{d^2y}{dx^2}\) converts into D(D - 1)y & x3\(\frac{d^3y}{dx^3}\) converts into D(D - 1)(D - 2)y. Differential equation (1) converts into D(D - 1)(D - 2)y + 3D(D - 1)y + Dy + y = ez + z ⇒ (D3 - 3D2 + 2D + 3D2 - 3D + D + 1)y = ez + z ⇒ (D3 + 1)y = ez + z---(2) It's auxiliarly equation is m3 + 1 = 0 ⇒ (m + 1) (m2 - m + 1) = 0 ⇒ m = -1, \(\frac{1\pm\sqrt3i}{2}\) \(\therefore\) C.F. = C1e-z + ez/2(C2 cos(\(\frac{\sqrt3z}2\)) + C3 sin(\(\frac{\sqrt3z}2\))) P.I. = \(\frac1{D^3+1}(e^z+z)\) = \(\frac1{D^3+1}e^z+\frac1{D^3+1}z\) = \(\frac{e^z}{1^3+11}+(1+D^3)^{-1}z\) = \(\frac{e^z}2+(1-D^3+....)z\) = \(\frac{e^z}2+z\) \(\therefore\) solution of differential equation (2) is y = C. F. + P. I. = C1e-z + (ez)1/2(C2 cos(\(\frac{\sqrt3}2\) log x) + C3 sin(\(\frac{\sqrt3}2\) log x)) + x/2 + log x which is complete solution of given differential equation. |
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