Question

If y = 6(x+1)+(a + bx)e^3x then show that y" -6y' +9y = 54x +18.....​

Answer 1

Given : y = 6(x+1)+(a + bx)e^3x

To Find : show that y" -6y' +9y = 54x +18

Solution:

y = 6(x + 1) + (a + bx)e³ˣ

=> y'  = 6  + (a + bx)3e³ˣ  + be³ˣ

y''  =   (a + bx)9e³ˣ    + 3be³ˣ     +  3be³ˣ

y" -6y' +9y

=  (a + bx)9e³ˣ  + 3be³ˣ  +  3be³ˣ - 6(6  + (a + bx)3e³ˣ  + be³ˣ)  + 9(6(x + 1) + (a + bx)e³ˣ)

=  (a + bx)9e³ˣ  + 6be³ˣ  - 36 - 18(a + bx)e³ˣ - 6be³ˣ + 54(x + 1) + 9(a + bx)e³ˣ

= (a + bx)e³ˣ(9 - 18 + 9)  + be³ˣ(6 - 6) -36 +  54x  + 54

= 0 + 0  - 36 + 54x + 54

= 54x + 18

Hence y" -6y' +9y = 54x + 18

QED

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