Find the general solution of 2y'' + 5y' = 100xe^(-x)cos x

Find the general solution of $2y'' + 5y' = 100xe^{-x}\cos x$.

Step 1: Homogeneous solution

\[ 2r^2+5r=0 \implies r(2r+5)=0 \implies r_1=0,\ r_2=-\frac52. \]

\[ Y_h = C_1 + C_2e^{-5x/2}. \]

Step 2: Normalize and set up the trial

Dividing the equation by $2$ puts it in standard form: $y''+\dfrac52 y' = 50xe^{-x}\cos x$. The right-hand side is $e^{\alpha x}P_1(x)\cos\beta x$ with $\alpha=-1,\ \beta=1$. The associated exponents $r=-1\pm i$ are checked against the roots $\{0,-5/2\}$: not a match, so $s=0$ and the full linear trial is kept:

\[ y^*=e^{-x}\bigl[(A_1x+A_0)\cos x+(B_1x+B_0)\sin x\bigr] = e^{-x}v. \]

Step 3: Exponential shift

Shifting $D\to D-1$ in $L(D)=D^2+\dfrac52D$: \[L(D-1) = (D-1)^2+\frac52(D-1) = D^2-2D+1+\frac52D-\frac52 = D^2+\frac12D-\frac32,\] so $v$ must satisfy $v''+\dfrac12v'-\dfrac32v = 50x\cos x$, using the same differentiated forms as in Problem 143: $v' = (A_1+B_1x+B_0)\cos x + (B_1-A_1x-A_0)\sin x$, $v'' = (2B_1-A_1x-A_0)\cos x + (-2A_1-B_1x-B_0)\sin x$.

Step 4: Match coefficients

Collecting $x\cos x,\ x\sin x,\ \cos x,\ \sin x$ in $v''+\dfrac12v'-\dfrac32v=50x\cos x$ gives the system: \[x\cos x:\ -\frac52A_1+\frac12B_1=50 \qquad x\sin x:\ -\frac12A_1-\frac52B_1=0\]

\[ \cos x:\ \frac12A_1+2B_1-\frac52A_0+\frac12B_0=0 \qquad \sin x:\ -2A_1+\frac12B_1-\frac12A_0-\frac52B_0=0 \]

From the second equation, $A_1=-5B_1$. Substituting into the first: $\dfrac{25}{2}B_1+\dfrac12B_1=50 \Rightarrow 13B_1=50 \Rightarrow B_1=\dfrac{50}{13},\ A_1=-\dfrac{250}{13}$.

Substituting these into the $\cos x$ and $\sin x$ equations and solving the resulting $2\times2$ system for $A_0,B_0$ (the bookkeeping is routine fraction arithmetic, carried out over the common denominator $169=13^2$) gives:

\[ A_0=\frac{400}{169}, \qquad B_0=\frac{2650}{169}. \]

Step 5: Assemble

Collecting all four coefficients over the denominator $169$ (note $-\frac{250}{13}=-\frac{3250}{169}$ and $\frac{50}{13}=\frac{650}{169}$), a common factor of $50$ emerges:

\[ v = \left(-\frac{250}{13}x+\frac{400}{169}\right)\cos x + \left(\frac{50}{13}x+\frac{2650}{169}\right)\sin x = \frac{50}{169}\Bigl[(-65x+8)\cos x+(13x+53)\sin x\Bigr]. \]

\[ y^* = e^{-x}v = \frac{50\,e^{-x}}{169}\Bigl[(-65x+8)\cos x+(13x+53)\sin x\Bigr]. \]

Direct substitution confirms this satisfies $2y''+5y'=100xe^{-x}\cos x$.

Step 6: General solution

The solution is recorded in the result below.