Question #289449

Find the integral curves of the differential equation:


(D³-D'³) z=x³y³

1
Expert's answer
2022-01-25T14:51:02-0500

The given equation can be written as

(D3D3)z=x3y3(D^3 – D'^3)z = x^3y^3 --------(1)

Corresponding auxiliary equation is

m31=0m^3 – 1 = 0 , i.e (m1)(m2+m+1)=0(m – 1)(m^2 + m + 1) = 0

or

(m1)(mw)(mw2)=0,(m – 1)(m – w)(m – w^2) = 0, for w to be the cube root of unity----(2)

C.F.=φ1(y+x)+φ2(y+wx)+φ3(y+w2x)\displaystyle C.F. = φ_{1}(y + x) + φ_2(y + wx) + φ_3(y + w^2x) -----(3)

Now

p.I=1D3D3x3y3=1D3[1(DD)3]x3y3\displaystyle p.I=\frac{1}{D^3-D^{\prime3}}x^3y^3=\frac{1}{D^3\left[1-(\frac{D\prime}{D})^3\right]}x^3y^3


=1D3[1(DD)3]1x3y3=1D3[1+D3D+...]x3y3=\frac{1}{D^3}\left[1-(\frac{D\prime}{D})^3\right]^{-1}x^3y^3=\frac{1}{D^3}\left[1+\frac{D^\prime3}{D}+...\right]x^3y^3


=1D3x3y3+D3D6x3y3=\frac{1}{D^3}x^3y^3+\frac{D^\prime3}{D^6}x^3y^3


(on taking DD\frac{D^\prime}{D}  to the powers to which y is appearing in

F(x,y)=xnymF(x, y) = x^ny^m i.e neglecting terms containing powers more than 3 in DD^\prime

=x6y3456+6x9456789\displaystyle =\frac{x^6y^3}{4\cdot5\cdot6}+\frac{6x^9}{4\cdot5\cdot6\cdot7\cdot8\cdot9}


P.I=x6y3120+x910080P.I\displaystyle =\frac{x^6y^3}{ 120}+\frac{x^9}{ 10080}

Hence complete solution 

z=φ1(y+x)+φ2(y+wx)+φ3(y+w2x)+x6y3120+x910080\displaystyle z=φ_{1}(y+x)+φ_{2}(y+wx)+φ_{3}(y+w^2x)+\frac{x^6y^3}{ 120}+\frac{x^9}{ 10080}


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