Question

(D^2-6DD'+9D'^2)z=12x^2+36xy​

Answer 1

Answer:

Step-by-step explanation:

From the above question,  

Given differential equation is (D^2-6DD'+9D'^2)z=12x^2+36xy

where D =$\frac{d}{dx},$

           D' = $\frac{d}{dy}$

Here we need to solve the given equation and find the differential  equation.

(D−3D ′) ^2 z=12x ^2 +36xy

So, Auxiliary equation is (D-3D')^2 = 0

 D - 3D' = 0,

 D − 3D' = 0

Hence, Complementary function (C.F.) = $\phi_1(y+3x) + x \phi_$2(y+3x)

Particular Integral (P.I.) =  $\frac{1}{(D-3D')^2} (12x^2 + 36 xy)$

$= \frac{1}{D^2} \left(1-\frac{3D'}D{}\right)^{-2} (12x^2+36xy)\\= \ \left(1+2\frac{3D'}{D}+3\right(\frac{3D'}{D}\left)^2+4\right(\frac{3D'}{D}\left)^3+…\right) \int\int(12x^2+36xy) dxdx$

$= x^4 + 6x^3 y + 9 x^4 \\ = 10 x^4 + 6x^3 y=x 4 +6x 3 y+9x 4 =10x 4 +6x 3 y$

Hence, solution of given differential equation is

z = C.F. + P.I.

$z=ϕ (y+3x)+xϕ (y+3x)+10x 4 +6x 3 y$

Hence, z = ϕ1(y + 3x) + xϕ2 (y + 3x) +10x^3 +6x^3y

solution of given differential equation is  ϕ1(y + 3x) + xϕ2 (y + 3x) +10x^3 +6x^3y

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