Question

solve : (D^3 + 2D^2 - D - 2)y = 1 - 4x^3​

Answer 1

The complete solution of the DE will be,

y = C.F. + P.I. = C1$e^{x}$ + C2$e^{- x}$ + C3$e^{- 2x}$  +  $2x^{3} - 3x^{2} + 15x - 8$

Given that;

(D^3 + 2D^2 - D - 2)y = 1 - 4x^3​

To find;

The solution of (D^3 + 2D^2 - D - 2)y = 1 - 4x^3​

Solution;

Auxillary equation  is $m^{3} + 2m^{2} - m^{} - 2$ = 0  

i.e., (m + 2) (m – 1)(m + 1) = 0

So that m = ± 1, – 2

Hence C.F. is ,

C.F. = C1$e^{x}$ + C2$e^{- x}$ + C3$e^{- 2x}$

Now we will find out the P.I. of the equation.

f(x) = 1 - $4x^{3}$

Then as f(x) is a polynomial of degree 3 we would look for a polynomial solution of the same degree, i.e. of the form:

y = $ax^{3} + bx^{2} + cx^{} + d$

Where the constants a,b,c and d is to be determined by direct substitution.

Differentiating y w.r.t x we get

y' = 3a$x^{2}$ + 2bx + c

Again differentiating w.r.t x ,

y" = 6ax + 2b

Again differentiating w.r.t x ,

y"' = 6a

Substituting these results into the DE [A] we get,

a = 2, b = - 3, c = 15 and d = - 8. Therefore,

The P.I. of the DE is ,

P.I. = $2x^{3} - 3x^{2} + 15x - 8$.

Hence, the complete solution of the DE will be,

y = C.F. + P.I. = C1$e^{x}$ + C2$e^{- x}$ + C3$e^{- 2x}$  +  $2x^{3} - 3x^{2} + 15x - 8$.

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Answer 2

Answer:

Step-by-step explanation:

From the above question,

They have given :

(D^3 + 2D^2 - D - 2)y = 1 - 4x^3​

Here we have to find,

m3 + 2m2 – m – 2 = 0

 i.e., (m + 2) (m – 1)(m + 1) = 0  

So that m = ± 1, – 2

Hence C.F. is C.F. = C1 ex + C2e–x + C3 e–2x  

Note that ex is common in C.F. and the R.H.S. of the given equation.  Therefore P.I. is of the form yp = a + bx + cx2 + dxex

We have to find a, b, c and d such that  y′′′p + 2y"p - y'p - 2yp = x2 + ex

y' = 3a + 2bx + c

Again differentiating w.r.t x ,

y" = 6ax + 2b

Again differentiating w.r.t x ,

y"' = 6a

Substituting these results into the DE [A] we get,

a = 2, b = - 3, c = 15 and d = - 8. Therefore,

The P.I. of the DE is ,

Hence, the complete solution of the DE will be,

y = C.F. + P.I. = C1 + C2 + C3  

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