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Solve X 2 D 3 Y D X 3 + 3 X D 2 Y D X 2 + D Y D X + Y X = 4 Log X - Applied Mathematics 2

Solve`  x^2 (d^3y)/dx^3+3x (d^2y)/dx^2+dy/dx+y/x=4log x` 


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`x^2 (d^3y)/dx^3+3x (d^2y)/dx^2+dy/dx+y/x=4log x`   

The given diff. eqn is Cauchy’s homogeneous eqn .
Multiply the given eqn by x, 

`x^3 (d^3y)/dx^3+3x^2 (d^2y)/dx^2+x dy/dx+y=4 x log x`

Put x = 𝒆𝒛 log x = z 

Diff. w.r.t x, 

`1/x=dz/dx`              but  `dy/dx=dy/ 1/x` 

∴ `x dy/dx=dy` 

`x^2 (d^2 y)/dx^2=D(D-1)y` 

`x^3 (d^3 y) /dx^3= D(D-1) (D-2)y` 

∴ `[D(D-1)(D-2)+3D(D-1)+D+1]y=4z.e^z`

∴ `[D^3+1]y=4z.e^z` 

For complementary solution , 


∴` [D^3+1]y=4z.e^z`

Roots are: `D=-1,1/2+isqrt3/2,1/2-isqrt3/2`   

Roots of the eqn are real and complex. 

∴` y_c=c_1e^-z+e^(z/2) (c_2cos sqrt(3z)/2+c_3 sin sqrt(3z)/2)` 

For particular integral , 

`y_p=1/f(D) x=1/(D^3+1) 4z.e^z` 

= `4e^z 1/((D+1)^3+1) Z` 

= `4e^Z 1/(D^3+3D^2+3D+2) z`

∴` y_p=e^z(2z-3)` 

The general solution of given diff. eqn is , 

`y_g=y_c+y_p=c_1e_(-z/2)(c_2cos  sqrt(3z)/2 + c_3 sin sqrt(3z)/2)+e^z(2z-3)`

Resubstitute z, 

∴ `y_g=c_1/x + sqrtx(c_2cos  sqrt(3 log x)/2+c_3 sin sqrt(3log x)/2)+(2 log x-3)` 

Concept: Linear Differential Equation with Constant Coefficient‐ Complementary Function
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