Answer to Question #209768 in Differential Equations for Nshalati

Question

(D²-2D+3)y=x²-1

Answer 1

$(D^2 - 2D +3) y = x^2 - 1$

The differential equation can be written as

$y'' - 2y' +3y = x^2 -1$

The auxiliary equation of the homogenous part is

$m^2-2m+3 = 0$

Solving the quadratic equation, we have that

$m = 1 \pm \sqrt{2}i$

Hence the solution of the homogenous part is

$y = e^x(c_1\cos \sqrt{2}x + i \,c_2\sin \sqrt{2}x)$

We use the method of undetermined coefficient to solve the other part.

Suppose

$y = ax^2 + bx + c$

Then

$y' = 2ax + b$

$y'' = 2a$

Substituting into the given differential equation, we have

$2a-2(2ax+b)+3(ax^2+bx+c) = x^2 - 1$

Comparing the coefficients of $x^2$ we have

$3a = 1 \implies a = \dfrac{1}{3}$

Comparing the coefficients of x, we have

$-4a +3b = 0 \implies b = \dfrac{4}{9}$

Comparing the constants, we have$2a -2b+3c = -1 \implies c = - \dfrac{7}{27}$

Hence, we have that the solution for this part is $y= \frac{1}{3}x^2 + \frac{4}{9}x - \frac{7}{27}$

So the general solution is

$y = e^x(c_1\cos \sqrt{2}x + i \,c_2\sin \sqrt{2}x) +\frac{1}{3}x^2 + \frac{4}{9}x - \frac{7}{27}$