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Evaluate ( 2 X + 1 ) 2 D 2 Y D X 2 − 2 ( 2 X + 1 ) D Y D X − 12 Y = 6 X - Applied Mathematics 2

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Sum

Evaluate `(2x+1)^2(d^2y)/(dx^2)-2(2x+1)(dy)/(dx)-12y=6x`

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Solution

`(2x+1)^2(d^2y)/(dx^2)-2(2x+1)(dy)/(dx)-12y=6x`

Put `(2x+1)=e^z  =>  x=(e^x-1)/2`

`(dz)/(dx)=2/(2x+1)`    but`(dy)/(dx)=(dy)/(dx)(dz)/(dx)=2(dy)/(dx)=2/(2x+1)"Dy"    "where"   "D"=d/(dz)`

`therefore(2x+1)(dy)/(dx)=2"Dy"`

`therefore(2x+1)^2(d^2y)/(dx^2)=2^2"D(D-1)y"`

From (1),

`4D(D-1)y-4Dy-12y=6((e^x-1)/2)`

`(4D^2-8D-12)y=3(e^z-1)`

For complementary solution ,

`(4D^2-8D-12)=0`

∴D = -1,3

`thereforey_c=c_1e^(-z)+c_2 e^(3z)`

For particular integral ,

`y_p=1/(f(D))X`

`y_p=1/(4D^2-8D-12)(3(e^z-1))`

`therefore y_p=3/4 1/(D^2-2D-3)(e^z-1)`    put D = a = 1 and D = a = 0

`thereforey_p=3/4(1/3-e^z/4)`

The general solution of given differential eqn is ,

`thereforey_g=y_c+y_p=c_1e^(-z)+c_2e^(3z)+3/4(1/3-e^z/4)`
Resubstituting 𝒛 ,

`therefore y_g=c_1(2x+1)^(-1)+c_2(2x+1)^3+3/4(1/3-(2x+1)/4)`

Concept: Linear Differential Equation with Constant Coefficient‐ Complementary Function
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