Solve (D^2-3D+2)y =x(x+4) and show that its general solution is given by y=Ae^x + Be^2x +(x^2/2) + (7x/2) + (19/4)
My work :
auxiliary eqn ,
m^2 - 3m + 2= 0
m=1 , m=2
Reduced eqn :
yn = Ae^x + Be^2x
P.I. = 1/f(D)*F(x) = (1/(D^2-3D+2))*(x(x+4))
P.I. = [(x^2+4x) + ( (1/2)*( 2- 3(2x) - 3(4) ) ) + (1/4)(9*2) )
P.I. = x^2 + (4x/2) - (6x/4) - (5/2) + (18/4)
P.I. =(x^2/2) + x + 8/4
General solution :
y= Ae^x + Be^2x + (x^2/2) + x + 8/4
Could anyone point out my mistakes?
Well, your final mistake was not checking to make sure your general solution worked.
to find one solution, set
u = 1/(D-2) x(x+4)
(D-2) u = x^2+4x
Now use the integrating factor e^(-2x) to get
(e^(-2x) u)' = (x^2+4x) e^(-2x)
e^(-2x) u = ∫ (x^2+4x) e^(-2x) dx = -1/4 e^(-2x) (2x^2+10x+5)
u = -1/4 (2x^2+10x+5)
Now, y = 1/(D-1) u
follow that through and you wind up with
y = Ae^x + Be^2x + 1/4 (2x^2 + 14x + 19)
Or, you can decompose 1/f(D) using partial fractions.
Is this a compulsory step for all D operator questions, checking whether the general solution works.
it ought to be compulsory when solving any math problem!
The fact that you arrive at an answer does not mean it is correct. We all make mistakes.
Your work and steps are correct up until the calculation of the particular integral (P.I) term.
To find the particular integral, you correctly stated that P.I = 1/f(D) * F(x), where f(D) is the characteristic equation (D^2-3D+2) and F(x) is the function to be solved (x(x+4)).
However, there seems to be a mistake in your calculation of the P.I. The expression you obtained is not correct.
Let's calculate the particular integral correctly:
Substitute D=1 and D=2 into the characteristic equation:
f(1) = (1)^2 - 3(1) + 2 = 1 - 3 + 2 = 0
f(2) = (2)^2 - 3(2) + 2 = 4 - 6 + 2 = 0
Since the roots are repeated, we need to use a different method to find the particular integral.
We use the method of "Annihilator" to solve this second-order linear differential equation.
1. Find the left-hand side of the equation:
D^2 - 3D + 2
2. Construct the annihilator by taking the product of (D-1)^2 and (D-2)^2:
(A^2)(D-1)^2(D-2)^2
3. Multiply the annihilator by the right-hand side of the equation:
(A^2)(D-1)^2(D-2)^2[x(x+4)]
4. Simplify the expression:
(A^2)(D^4 - 6D^3 + 9D^2 - 4D + 4)[x(x+4)]
5. Expand the expression:
(A^2)[x^2(D^4 - 6D^3 + 9D^2 - 4D + 4) + 4x(D^4 - 6D^3 + 9D^2 - 4D + 4)]
6. Collect like terms and simplify:
(A^2)[(1+4)x^2(D^4 - 6D^3 + 9D^2 - 4D + 4) + 4x(D^4 - 6D^3 + 9D^2 - 4D + 4)]
7. Combine similar terms:
(A^2)[(x^2+4x)(D^4 - 6D^3 + 9D^2 - 4D + 4)]
Now, divide both sides of the equation by the constant coefficient (D^4 - 6D^3 + 9D^2 - 4D + 4) to isolate the particular integral term on the left side.
P.I. = (x^2+4x)/(D^4 - 6D^3 + 9D^2 - 4D + 4)
The particular integral term simplifies to:
P.I. = (x^2+4x)/(D-1)^2(D-2)^2
Now, you can substitute this particular integral term back into the general solution and solve for the constants A and B to obtain the final solution.