Given ( D 2 + 5 D D ′ + 5 D ′ 2 ) z = x . s i n ( 3 x − 2 y ) (D^2+5DD'+5D'^2)z= x.sin(3x-2y) ( D 2 + 5 D D ′ + 5 D ′2 ) z = x . s in ( 3 x − 2 y )
The auxiliary equation is
m 2 + 5 m + 5 = 0 ⟹ − b ± b 2 − 4 a c 2 a = − 5 ± 5 2 − 4 ( 1 ) ( 5 ) 2 ( 1 ) i . e . m = − 5 , 0 m^2+5m+5=0 \implies \frac{-b \pm \sqrt{b^2-4ac}}{2a} =\frac{-5 \pm \sqrt{5^2-4(1)(5)}}{2(1)} \\
i.e.\ m=-5,0\\ m 2 + 5 m + 5 = 0 ⟹ 2 a − b ± b 2 − 4 a c = 2 ( 1 ) − 5 ± 5 2 − 4 ( 1 ) ( 5 ) i . e . m = − 5 , 0
whence complementary function (C.F.) =
ϕ 1 ( y − 5 x ) + ϕ 2 ( y ) \phi_1(y-5x)+\phi_2(y)\\ ϕ 1 ( y − 5 x ) + ϕ 2 ( y ) P . I . = 1 D 2 + 5 D D ′ + 5 D ′ 2 x ⋅ s i n ( 3 x − 2 y ) = { D 2 = − 9 D D ′ = 6 D ′ = − 4 P.I. = \frac{1}{D^2+5DD'+5D'^2}x \cdot sin(3x-2y)=
\begin{cases}
D^2=-9 \\
DD'=6 \\
D'=-4
\end{cases} P . I . = D 2 + 5 D D ′ + 5 D ′2 1 x ⋅ s in ( 3 x − 2 y ) = ⎩ ⎨ ⎧ D 2 = − 9 D D ′ = 6 D ′ = − 4
(i.e., on replacing D 2 D^2 D 2 a 2 , D D ′ a^2,\ DD' a 2 , D D ′ − a b , D ′ 2 = − b 2 -ab, D'^2=-b^2 − ab , D ′2 = − b 2 φ ( – a 2 , – a b , – b 2 ) ≠ 0 φ(–a2, –ab, –b2) ≠ 0 φ ( – a 2 , – ab , – b 2 ) = 0
Clearly it is a case of failure as φ ( – a 2 , – a b , – b 2 ) = 0. φ(–a2, –ab, –b2) = 0. φ ( – a 2 , – ab , – b 2 ) = 0.
Therefore, factorize ϕ ( D , D ′ ) \phi (D,D') ϕ ( D , D ′ )
P . I . = 1 D ( D + 5 D ′ ) x s i n ( 3 x − 2 y ) = 1 D ∫ D + 5 D ′ x s i n ( 3 x − 10 x − 2 c 1 ) ∴ y − 5 x = c 1 = 1 D ∫ x s i n ( − 7 x − 2 c 1 ) δ x = 1 D ∫ − x c o s ( − 7 x − 2 c 1 ) − 7 + 1 − 7 s i n ( − 7 x − 2 c 1 ) − 7 = 1 D x 7 c o s ( 3 x − 2 y ) + 1 49 s i n ( 3 x − 2 y ) = ∫ D ( x 7 c o s ( 3 x − 2 y ) + 1 49 s i n ( 3 x − 2 y ) ) δ x = ∫ D ( x 7 c o s ( 3 x − 2 c 2 ) + 1 49 s i n ( 3 x − 2 c 2 ) ) δ x ∴ y = c 2 P.I.=\frac{1}{D(D+5D')}x\ sin (3x-2y)\\
=\frac{1}{D} \intop_{D+5D'}x\ sin (3x-10x-2c_1)\ \therefore\ y-5x=c_1\\
=\frac{1}{D} \intop x\ sin (-7x-2c_1)\delta x =\frac{1}{D} \intop \frac{- x\ cos (-7x-2c_1)}{-7}+\frac{1}{-7}\frac{sin (-7x-2c_1)}{-7}\\
=\frac{1}{D}\frac{x}{7}cos(3x-2y)+\frac{1}{49}sin (3x-2y)\\
= \intop_D(\frac{x}{7}cos(3x-2y)+\frac{1}{49}sin(3x-2y))\delta x\\
= \intop_D(\frac{x}{7}cos(3x-2c_2)+\frac{1}{49}sin(3x-2c_2))\delta x\ \therefore y=c_2\\ P . I . = D ( D + 5 D ′ ) 1 x s in ( 3 x − 2 y ) = D 1 ∫ D + 5 D ′ x s in ( 3 x − 10 x − 2 c 1 ) ∴ y − 5 x = c 1 = D 1 ∫ x s in ( − 7 x − 2 c 1 ) δ x = D 1 ∫ − 7 − x cos ( − 7 x − 2 c 1 ) + − 7 1 − 7 s in ( − 7 x − 2 c 1 ) = D 1 7 x cos ( 3 x − 2 y ) + 49 1 s in ( 3 x − 2 y ) = ∫ D ( 7 x cos ( 3 x − 2 y ) + 49 1 s in ( 3 x − 2 y )) δ x = ∫ D ( 7 x cos ( 3 x − 2 c 2 ) + 49 1 s in ( 3 x − 2 c 2 )) δ x ∴ y = c 2
= \intop(\frac{x}{7}cos(3x-2c_2)\delta x+\intop\frac{1}{49}sin(3x-2c_2))\delta x\
Letting a = ∫ ( x 7 c o s ( 3 x − 2 c 2 ) δ x a= \intop(\frac{x}{7}cos(3x-2c_2)\delta x a = ∫ ( 7 x cos ( 3 x − 2 c 2 ) δ x b=\intop\frac{1}{49}sin(3x-2c_2))\delta x\
a = ∫ ( x 7 c o s ( 3 x − 2 c 2 ) δ x = x s i n ( 3 x − 2 c 2 ) 21 + 1 3 c o s ( 3 x − 2 c 2 ) 3 a ⟹ x 21 s i n ( 3 x − 2 c 2 ) + 1 9 c o s ( 3 x − 2 c 2 ) b = ∫ ( 1 49 s i n ( 3 x − 2 c 2 ) δ x = − c o s ( 3 x − 2 c 2 ) 49 + 1 3 s i n ( 3 x − 2 c 2 ) 49 b ⟹ − 1 49 c o s ( 3 x − 2 c 2 ) + 1 147 s i n ( 3 x − 2 c 2 ) ∴ a + b ⟹ x 21 s i n ( 3 x − 2 c 2 ) + 1 9 c o s ( 3 x − 2 c 2 ) − 1 49 c o s ( 3 x − 2 c 2 ) + 1 147 s i n ( 3 x − 2 c 2 ) ⟹ 7 x 147 s i n ( 3 x − 2 c 2 ) + 1 147 s i n ( 3 x − 2 c 2 ) + 40 441 c o s ( 3 x − 2 c 2 ) P . I . = 7 x 147 s i n ( 3 x − 2 y ) + 1 147 s i n ( 3 x − 2 y ) + 40 441 c o s ( 3 x − 2 y ) a= \intop(\frac{x}{7}cos(3x-2c_2)\delta x =\frac{x\ sin(3x-2c_2)}{21}+\frac13 \frac{cos(3x-2c_2)}{3}\\
a\implies \frac{x}{21}sin(3x-2c_2) +\frac{1}{9}cos(3x-2c_2)\\
b= \intop(\frac{1}{49}sin(3x-2c_2)\delta x =\frac{-cos(3x-2c_2)}{49}+\frac13 \frac{sin(3x-2c_2)}{49}\\
b\implies -\frac{1}{49}cos(3x-2c_2) +\frac{1}{147}sin(3x-2c_2)\\
\therefore a+b \implies \frac{x}{21}sin(3x-2c_2) +\frac{1}{9}cos(3x-2c_2)-\frac{1}{49}cos(3x-2c_2) +\frac{1}{147}sin(3x-2c_2)\\
\implies \frac{7x}{147}sin(3x-2c_2)+\frac{1}{147}sin(3x-2c_2) +\frac{40}{441}cos(3x-2c_2)\\
P.I. =\frac{7x}{147}sin(3x-2y)+\frac{1}{147}sin(3x-2y) +\frac{40}{441}cos(3x-2y)\\ a = ∫ ( 7 x cos ( 3 x − 2 c 2 ) δ x = 21 x s in ( 3 x − 2 c 2 ) + 3 1 3 cos ( 3 x − 2 c 2 ) a ⟹ 21 x s in ( 3 x − 2 c 2 ) + 9 1 cos ( 3 x − 2 c 2 ) b = ∫ ( 49 1 s in ( 3 x − 2 c 2 ) δ x = 49 − cos ( 3 x − 2 c 2 ) + 3 1 49 s in ( 3 x − 2 c 2 ) b ⟹ − 49 1 cos ( 3 x − 2 c 2 ) + 147 1 s in ( 3 x − 2 c 2 ) ∴ a + b ⟹ 21 x s in ( 3 x − 2 c 2 ) + 9 1 cos ( 3 x − 2 c 2 ) − 49 1 cos ( 3 x − 2 c 2 ) + 147 1 s in ( 3 x − 2 c 2 ) ⟹ 147 7 x s in ( 3 x − 2 c 2 ) + 147 1 s in ( 3 x − 2 c 2 ) + 441 40 cos ( 3 x − 2 c 2 ) P . I . = 147 7 x s in ( 3 x − 2 y ) + 147 1 s in ( 3 x − 2 y ) + 441 40 cos ( 3 x − 2 y )
∴ \therefore ∴ z = C . I . + P . I . z=C.I.+P.I. z = C . I . + P . I .
Answer
= ϕ 1 ( y − 5 x ) + ϕ 2 ( y ) + 7 x 147 s i n ( 3 x − 2 y ) + 1 147 s i n ( 3 x − 2 y ) + 40 441 c o s ( 3 x − 2 y ) =\phi_1(y-5x)+\phi_2(y)+\frac{7x}{147}sin(3x-2y)+\frac{1}{147}sin(3x-2y) +\frac{40}{441}cos(3x-2y)\\ = ϕ 1 ( y − 5 x ) + ϕ 2 ( y ) + 147 7 x s in ( 3 x − 2 y ) + 147 1 s in ( 3 x − 2 y ) + 441 40 cos ( 3 x − 2 y )
#340153 on Dec 2023
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