solve (D2+3D+2)y=1+3x+x2

Question

it is from differential equations 2 from maths2 ksc

Answer 1

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.