Find the general solution of the given differential equation: $\displaystyle y^{6} - 3y^{4} + 3y'' - y = 0 $
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Originally Posted by fudawala Find the general solution of the given differential equation: $\displaystyle y^{6} - 3y^{4} + 3y'' - y = 0 $ $\displaystyle k^6 - 3k^4 + 3k^2 - 1 = 0$ $\displaystyle (k^2-1)^3 = 0$ $\displaystyle (k-1)(k-1)(k-1)(k+1)(k+1)(k+1)=0$ $\displaystyle y=Ae^x+Bxe^x+Cx^2e^x+De^{-x}+Exe^{-x}+Fx^2e^{-x}$
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