Last Activity: 4 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)
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
AskIItians Expert
Prepraring for the competition made easy just by live online class.
Full Live Access
Study Material
Live Doubts Solving
Daily Class Assignments
Get your questions answered by the expert for free
Last Activity: 2 Years ago
Last Activity: 2 Years ago
Last Activity: 2 Years ago
Last Activity: 2 Years ago
Last Activity: 2 Years ago