Answer to Question #287366 in Differential Equations for amulya

Question

1. (2x+1)^2y''-2(2x+1)y'-12y=xlog(2x+1)
2. (x+2)^2y''-(x+2)y'+y=3x+4 these two r from legendres

Answer 1

$1)\\ (2x+1)^2y''-2(2x+1)y'-12y=xln(2x+1)\\ 2x+1=e^z,~~\\ \Rightarrow z=ln(2x+1),~~ x=\dfrac{e^z-1}{2}\\ \therefore z'=\dfrac{2}{2x+1}\\ \Rightarrow y'=Dy.z',~~~~~~D=\dfrac{d}{dz}\\ \Rightarrow (2x+1)y'=2Dy, ~~(2x+1)^2y''=4D(D-1)y\\~\\ \therefore 4D(D-1)y-4Dy-12y=z(\frac{e^z-1}{2}\\ \Rightarrow (4D^2-8D-12)y=ze^z/2-z/2\\ \text{Now, }\\ 4D^2-8D-12=0 \Rightarrow D=3,-1\\ \therefore CF=c_1e^{3z}+c_2e^{-z}\\$

$\\~\\ PI=\dfrac{ze^z-z}{2(4D^2-8D-12)}=\dfrac{ze^z-z}{8(D-3)(D+1)}\\ =\dfrac{1}{8(D-3)}\times e^{-z}\int(ze^z-z)e^z dz\\ =\dfrac{1}{8(D-3)}\times (ze^{2z}/2-e^{2z}/4-ze^z+e^z)\\ =\dfrac{1}{8}\times e^{3z}\int (ze^{2z}/2-e^{2z}/4-ze^z+e^z)e^{-3z}dz\\ =-\dfrac{1}{8}(\dfrac{ze^z}{4}-\dfrac{z}{3}+\dfrac{2}{9})$

Soln:-

$y=CF+PI=c_1e^{3z}+c_2e^{-z}-\dfrac{1}{8}(\dfrac{ze^z}{4}-\dfrac{z}{3}+\dfrac{2}{9})\\ =c_1(2x+1)^3+c_2(2x+1)^{-1}\\-\dfrac{1}{8}(\dfrac{ln(2x+1)(2x+1)}{4}-\dfrac{2x+1}{3}+\dfrac{2}{9})$

$2)\\ (x+2)^2y''-(x+2)y'+y=3x+4\\ x+2=e^z,~~\\ \Rightarrow z=ln(x+2),~~ x=e^z-2\\ \therefore z'=\dfrac{1}{x+2}\\ y'=Dy.z',~~~~~~D=\dfrac{d}{dz}\\ \Rightarrow (x+2)y'=Dy, ~~(x+2)^2y''=D(D-1)y\\~\\ \therefore D(D-1)y-Dy+y=3e^z-2\\ \Rightarrow (D^2-2D+1)y=3e^z-2\\ \text{Now, }\\ D^2-2D+1=0 \Rightarrow D=1\\ \therefore CF=ce^{z}$

$PI=\dfrac{3e^z-2}{(D-1)^2}=e^z\int(\int(3e^z-2)\times e^{-z}dz)dz\\ =\dfrac{3}{2}z^2e^z-2\\ ~~\\~~\\ \text{Solution :}\\ y=CF+PI=ce^z+\dfrac{3}{2}z^2e^z-2\\=-2+(c+\dfrac{3}{2}z^2)e^z\\ =(c+\dfrac{3}{2}(ln(x+2))^2)(x+2)-2$