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Find the particular integral of the given equation 1.(D^3-D^2-6D) y=1+x^2 2.(D^2+3D+2) y=1+3x+x^2

Find the particular integral of the given equation
1.(D^3-D^2-6D) y=1+x^2
2.(D^2+3D+2) y=1+3x+x^2
 

Grade:12th pass

1 Answers

Manika gupta
askIITians Faculty 43 Points
3 years ago

Dear Student
Find your solution here.

1)
Symbolic form becomes: (D3 – D2 – 6D)y = 1 + x2
Auxiliary Equation: D3 – D2 – 6D = 0
Implying D(D + 2)(D – 3) = 0 or D = 0, 3, – 2
yC.F. = (c1 + c2e3x + c3e– 2x)

631-57_Capture.PNG
2)
The “zeroes” of the left-hand side give us the exponent coefficients for the characteristic equation:

Ae⁻ˣ+Be⁻²ˣ where A and B are constants.

Now we look at the right-hand side and assume we can find a solution for y=ax²+bx+c where a, b and c are constants. Next we find values for the constants that will match 1+3x+x².

y'=2ax+b, y''=2a, so y''+3y'+2y=2a+6ax+3b+2ax²+2bx+2c=1+3x+x².

Matching coefficients we get:

x²: 2a=1, so a=½

x: 6a+2b=3, 3+2b=3, so b=0

constant: 2a+3b+2c=1, 1+0+2c=1, so c=0, making y=x²/2.

The complete solution is y=Ae⁻ˣ+Be⁻²ˣ+x²/2

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