Answer to Question #314933 in Differential Equations for msizi

Question

(D^2-2D+5)y=x+5

Answer 1

Solution

For the homogeneous equation the characteristic equation is

λ² - 2λ + 5=0  =>  $\lambda_{1,2}=1\pm\sqrt{1-5}$   => λ₁ = 1+2i, λ₂ = 1-2i

So the solution of homogeneous equation is

$y_0\left(x\right)=C_1e^xsin(2x)+C_2e^xcos(2x)$

where C₁, C₂ are arbitrary constants.

Partial solution can be found in the form y₁(x) = Ax+B. Substitution into equation gives

-2A+5(Ax+B)=x+5 => 5Ax = x, 5B - 2A = 5 => A = 0.2, B = 0.2*(5 + 0.4) = 1.08

Therefore the solution of the given equation is

${y(x)=y}_0\left(x\right)+y_1\left(x\right)=C_1e^xsin(2x)+C_2e^xcos(2x)+0.2x+1.08$