Question

(D²+D+1 )y= x² find particular integral​

Answer 1

Answer:

Answer

We have to solve (D

2

−4D+1)y=x

2

The characteristic equation is p

2

−4p+1=0

⇒p=

2

4±

16−4

=

2

4±2

3

=2±

3

Thus Complementary function C.F.=Ae

(2+

3

)x

+Be

(2−

3

)x

Particular integral P.I.=

D

2

−4D+1

1

(x

2

)

=[1−(4D−D

2

)]

−1

(x

2

)

=[1+(4D−D

2

)+(4D−D

2

)

2

+...](x

2

)

=[1+4D+15D

2

+...](x

2

)

∴P.I.=x

2

+8x+30

Hence the general solution is y=C.F.+P.I.

y.=Ae

(2+

3

)x

+Be

(2−

3

)x

+(x

2

+8x+30

Answer 2

The particular integral is  $\frac{x^2}{D^2 + D + 1}$]

Given:

(D²+D+1 )y= x²

To Find:

The particular integral of the given equation (D²+D+1 )y= x²

Solution:

The given equation can be written as,

$(m^{2} + m +1 )y = 0$

Complementary function,

m  $= \frac{-1 + 3i}{2}$ ; $\frac{-1-3i}{2}$

Particular Integral,

= $\frac{x^{2} }{D^{2} + D +1}$

General Solution = C.F + P.I

=  ${{\frac{-1 + 3i}{2}$ + $\frac{x^{2} }{D^{2} + D +1}$

#SPJ2