Question

Use Laplace transforms to solve the following initial value problem: x"+8x'+15x = 0; x(0) = 2, x'(0) = -3 PLEASE SHOW ALL WORK, OR RISK LOSING ALL POINTS!!!! x')=sX (s) - x(0) x"(t) = sº X(s) - sx(0) - x'(0)

Answer 1

Taking the transform of both sides gives

$\mathcal L_s\{x''+8x'+15x\}=0$

$(s^2X(s)-sx(0)-x'(0))+8(sX(s)-x(0))+15X(s)=0$

where $X(s)$ denotes the Laplace transform of $x(t)$, $\mathcal L_s\{x(t)\}$. Solve for $X(s)$ to get

$(s^2+8s+15)X(s)=2s+13$

$X(s)=\dfrac{2s+13}{s^2+8s+15}=\dfrac{2s+13}{(s+3)(s+5)}$

Split the right side into partial fractions:

$\dfrac{2s+13}{(s+3)(s+5)}=\dfrac a{s+3}+\dfrac b{s+5}$

$2s+13=a(s+5)+b(s+3)$

If $s=-3$, then $7=2a\implies a=\dfrac72$; if $s=-5$, then $3=-2b\implies b=-\dfrac32$. So

$X(s)=\dfrac72\dfrac1{s+3}-\dfrac32\dfrac1{s+5}$

Finally, take the inverse transform of both sides to solve for $x(t)$:

$x(t)=\dfrac72e^{-5t}-\dfrac32e^{-3t}$