# 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

Sum

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

#### 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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