Author Topic: Problem 3 (afternoon)  (Read 1640 times)

Ranran Wang

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Re: Problem 3 (afternoon)
« Reply #15 on: October 23, 2019, 11:55:19 AM »
\textbf{Q3. $y^{\prime \prime}-5 y^{\prime}+6 y=52 \cos (2 x)$,$y(0)=0$,$y^{\prime}(0)=0$}

\textbf{Ans:} let $y=e^{r x}$

$y^{\prime}=r e^{r x}$

$y^{\prime \prime}=r^{2} e^{r x}$

$r^{2} e^{r x}-5 r e^{r x}+6 e^{r x}=0$

$\left(r^{2}-5 r+6\right)=0$

$(r-2)(r-3)=0$

$r_{1}=2$

$r_{2}=3$

$y=a e^{r x}+c_{2} e^{r_{2} x}=u e^{2 x}+c_{2} e^{3 x}$

let $y=A \cos (2 x)+B \sin (2 x)$

$y^{\prime}=-2 A \sin (2 x)+2 B \cos (2 x)$

$y^{\prime \prime}=-4 A \cos (2 x)-4 B \sin (2 x)$

$y^{\prime \prime}-5 y^{\prime}+6 y=52 \cos (2 x)$

$-4 A \cos (2 x)-4 B \sin (2 x)-5(-2 A \sin (2 x)+2 B \cos (2 x))+6(A \cos (2 x)+B \sin (2 x))=52(\sin (2 x)$

$(-4 A-10 B+6 A) \cos (2 x)+(-4 B+10 A+6 B) \sin (2 x)=52 \cos (2 x)$

$(-10 B+2 A) \cos (2 x)+(10 A+2 B) \sin (2 x)=52 \cos (2 x)$

$\left\{\begin{array}{c}{-10 B+2 A=52} \\ {10 A+2 B=0}\end{array}\right.$ $\Rightarrow 5 A+B=0$

$\Rightarrow \quad B=-5 A$  Sub $B=-5 A$ into $-10 B+2 A=52$

$50 A+2 A=52$

$A=1$

$B=-5$

$\therefore y=\cos (2 x)-5 \sin (2 x)$

$\therefore y=c_{1} e^{2 x}+c_{2} e^{3 x}+\cos (2 x)-5 \sin (2 x)$

then $y^{\prime}=2 C_{1} e^{2 x}+3 C_{2} e^{3 x}-2 \sin (2 x)-10 \cos (2 x)$

$y(0)=0$  sub $x=0, \quad y=0$

$0=C_{1} e^{0}+\left(2 e^{0}+\cos (0)-5 \sin (0)\right)=C_{1}+C_{2}+1=0$

$C_{2}=-1-C_{1}$

$y^{\prime}(0)=0$  sub $x=0, y^{\prime}=0$
$0=2C_{1}  e^{0}+3\left(2 e^{0}-2 \sin (0)-10 \cos (0)=2 C_{1}+3 C_{2}-10=0\right.$

$2C_{1}+3 C_{2}=10$

Sub $c_{2}=-1-C_{1}$ info $2C_{1}+3 C_{2}=10$

$2C_{1}+\left(-3-3 C_{1}\right)=10$

$2 C_{1}-3-3 C_{1}=10$

$C_{1}=-13$

$C_{2}=-1+13=12$

$\therefore y=-13 e^{2 x}+12 e^{3 x}+\cos (2 x)-5 \sin (2 x)$