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Question Number 91036 by niroj last updated on 27/Apr/20

$$\mathrm{Solve}:$$$${(\mathrm{x}+\mathrm{a})}^2\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}-4(\mathrm{x}+\mathrm{a})\frac{\mathrm{dy}}{\mathrm{dx}}+6\mathrm{y}=\mathrm{x}$$

Answered by MWSuSon last updated on 27/Apr/20

$$letx+a=e^z\Rightarrow x=e^z-a$$$${(x+a)}^2\frac{d^{2}y}{{dx}^2}=D(D-1)y$$$$(x+a)\frac{dy}{dx}=Dy$$$$D(D-1)y-4Dy+6y=e^z-a$$$$(D^2-D-4D+6)y=e^z-a$$$$(D^2-5D+6)y=e^z-a$$$$Auxillaryeauation$$$$m^2-5m+6=0$$$$(m-2)(m-3)=0$$$$m=2,3$$$$y_c=C_1{e}^{2z}+C_2{e}^{3z}$$$$y_p=\frac{1}{D^2-5D+6}(e^z-a)$$$$=\frac{1}{D^2-5D+6}{e}^z-\frac{1}{D^2-5D+6}a$$$$=\frac{1}{1-5+6}{e}^z-a\frac{1}{D^2-5D+6}{e}^{0x}$$$$=\frac{e^z}{2}-a\frac{1}{0-0+6}{e}^{0x}$$$$=\frac{e^z}{2}-\frac{a}{6}$$$$y=C_1{e}^{2z}+C_2{e}^{3z}+\frac{e^z}{2}-\frac{a}{6}$$$$butx+a=e^z\Rightarrow z={log}_e(x+a)$$$$y=C_1{(x+a)}^2+C_2{(x+a)}^3+\frac{(x+a)}{2}+\frac{a}{6}$$

Commented by peter frank last updated on 27/Apr/20

$$thankyou$$

Commented by peter frank last updated on 27/Apr/20

$$thankyou$$

Commented by niroj last updated on 03/May/20

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