Answer to Question #154634 in Differential Equations for Jagan

Question

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

Answer 1

Given: $(D^2+2D+5)y=xe^x$

Here, $D=\frac{d}{dx}$

So, we can rewrite it as: $\frac{d^2y}{dx^2}+2\frac{dy}{dx}+5y=xe^x$ ... (i)

First, we find $y_c$ , complimentary function.

Consider $\frac{d^2y}{dx^2}+2\frac{dy}{dx}+5y=0$

Its auxiliary equation is: $m^2+2m+5=0$

On solving by quadratic formula,

$m=\frac{-2\pm\sqrt{2^2-4(1)(5)}}{2(1)}=\frac{-2\pm\sqrt{4-20}}{2}=-1\pm2i$

Thus, the solution of homogeneous equation is:

$y_c=e^{-1x}(A\cos2x+B\sin2x)=Ae^{-x}\cos2x+Be^{-x}\sin2x$

Now, we find particular solution of non-homogeneous equation.

Assume $y=(ax+b)e^x$ be the solution ...(ii)

We find a and b by substitution and comparison.

On differentiating (ii) w.r.t $x$ ,

$\frac{dy}{dx}=(ax+b)e^x+(a)e^x \\ \Rightarrow\frac{dy}{dx}=(ax+a+b)e^x$

Again, differentiating it w.r.t $x$ ,

$\frac{d^2y}{dx^2}=(ax+a+b)e^x+ae^x \\ \Rightarrow \frac{d^2y}{dx^2}=(ax+2a+b)e^x$

Now substituting these values in (i), we get,

$(ax+2a+b)e^x+2(ax+a+b)e^x+5(ax+b)e^x=xe^x \\ \Rightarrow (ax+2a+b)+2(ax+a+b)+5(ax+b)=x$

$\Rightarrow (8a)x+(4a+8b)=1x+0$

On comparing,

$8a=1\Rightarrow a=\frac 1 8 \\ 4a+8b=0 \Rightarrow 8b=-4a \Rightarrow b=-\frac{a}{2}$

So, $b=-\frac1 2 \times \frac 1 8=-\frac 1 {16}$

So, $y_p=(\frac1 8x-\frac 1 {16})e^x=\frac{xe^x}{8}-\frac{e^x}{16}$

Thus, $y(x)=y_c+y_p$

$=Ae^{-x}\cos2x+Be^{-x}\sin2x+\frac{xe^x}{8}-\frac{e^x}{16}$

Answer: $y(x)=Ae^{-x}\cos2x+Be^{-x}\sin2x+\frac{xe^x}{8}-\frac{e^x}{16}$