## MATH1013 Quizzes

Quiz 11: Second order linear differential equations with constant coefficients
Question 1 Questions
What is the auxiliary equation associated with the second order linear differential equation
$\frac{{d}^{2}x}{d{t}^{2}}+19\frac{dx}{dt}-27x=0\phantom{\rule{0.3em}{0ex}}?$
 a) $\frac{{d}^{2}y}{d{x}^{2}}+19\frac{dy}{dx}-27y=0$ b) $\frac{{d}^{2}y}{d{t}^{2}}+19\frac{dy}{dt}-27y=0$ c) ${k}^{2}+19k-27=0$ d) ${k}^{2}-19k+27=0$

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is correct!
Choice (d) is incorrect
Find the general solution of the second order linear differential equation
$\frac{{d}^{2}y}{d{x}^{2}}-\frac{dy}{dx}-12y=0.$
 a) $y=A{e}^{4x}+B{e}^{-3x}$ b) $y=A{e}^{-4x}+B{e}^{3x}$ c) $y=A{e}^{6x}+B{e}^{-2x}$ d) $y=A{e}^{-6x}+B{e}^{2x}$

Choice (a) is correct!
${k}^{2}-k-12=\left(k-4\right)\left(k+3\right)=0$ hence $k=4$ or $k=-3$ and the solution is $y=A{e}^{4x}+B{e}^{-3x}.$
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is incorrect
Find the general solution to the second order linear differential equation
$2\frac{{d}^{2}x}{d{t}^{2}}-8\frac{dx}{dt}+6x=0.$
 a) $x=A{e}^{\left(-4+\sqrt{10}\right)t}+B{e}^{\left(-4-\sqrt{10}\right)t}$ b) $x=A{e}^{\left(4+\sqrt{10}\right)t}+B{e}^{\left(4-\sqrt{10}\right)t}$ c) $x=A{e}^{3t}+B{e}^{t}$ d) $x=A{e}^{-3t}+B{e}^{-t}$

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is correct!
$2{k}^{2}-8k+6=2\left(k-3\right)\left(k-1\right)=0$ hence $k=3$ or $k=1$ and the solution is $x=A{e}^{3t}+B{e}^{t}$.
Choice (d) is incorrect
Find the particular solution to the second order linear differential equation
$\frac{{d}^{2}y}{d{t}^{2}}+6\frac{dy}{dt}+8y=0,$
where $y=-2$ and $\frac{dy}{dt}=10$ when $t=0\phantom{\rule{0.3em}{0ex}}.$
 a) $y=A{e}^{-2t}+B{e}^{-4t}$ b) $y=A{e}^{2t}+B{e}^{4t}$ c) $y=-9{e}^{2t}+7{e}^{4t}$ d) $y={e}^{-2t}-3{e}^{-4t}$

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is correct!
${k}^{2}+6k+8=\left(k+2\right)\left(k+4\right)=0$ hence $k=-2$ or $k=-4$ and the general solution is $y=A{e}^{-2t}+B{e}^{-4t}\phantom{\rule{0.3em}{0ex}}.$ Also $\frac{dy}{dt}=-2A{e}^{-2t}-4B{e}^{-4t}\phantom{\rule{0.3em}{0ex}}.$ At $t=0$ we have $A+B=-2$ and $-2A-4B=10$ which gives $A=1$ and $B=-3\phantom{\rule{0.3em}{0ex}}.$
Find the general solution to the second order linear differential equation
$\frac{{d}^{2}y}{d{x}^{2}}-14\frac{dy}{dx}+51y=0.$
 a) $y={e}^{7x}\left(Asin\sqrt{2}x+Bcos\sqrt{2}x\right)$ b) $y={e}^{\sqrt{2}x}\left(Asin7x+Bcos7x\right)$ c) $y=A{e}^{3x}+B{e}^{-17x}$ d) $y=A{e}^{-3x}+B{e}^{17x}$

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is correct!
${k}^{2}-14k-51=\left(k-17\right)\left(k+3\right)=0$ hence $k=17$ or $k=-3$ and the general solution is $y=A{e}^{-3x}+B{e}^{17x}\phantom{\rule{0.3em}{0ex}}.$
Find the particular solution to the second order linear differential equation
$\frac{{d}^{2}y}{d{x}^{2}}-6\frac{dy}{dx}+4y=0$
where $y=6$ and $\frac{dy}{dx}=18$ when $x=0$.
 a) $y=3{e}^{\left(3+\sqrt{5}\right)x}+3{e}^{\left(3-\sqrt{5}\right)x}$ b) $y=6{e}^{3x}cos\sqrt{5}x\phantom{\rule{0.3em}{0ex}}.$ c) $y={e}^{\sqrt{5}x}\left(\left(6-2\sqrt{5}\right)sin3x+6cos3x\right)\phantom{\rule{0.3em}{0ex}}.$ d) The initial conditions are inconsistent since the general solution is$y=A{e}^{\left(-3+\sqrt{5}\right)x}+B{e}^{\left(-3-\sqrt{5}\right)x}\phantom{\rule{0.3em}{0ex}}.$

Choice (a) is correct!
${k}^{2}-6k-16=0$ has solutions $k=3±\sqrt{5}$ hence the general solution is $y=A{e}^{\left(3+\sqrt{5}\right)x}+B{e}^{\left(3-\sqrt{5}\right)x}\phantom{\rule{0.3em}{0ex}}.$ Since $y=6$ when $x=0$ we have $A+B=6$. Now
$\frac{dy}{dx}=\left(3+\sqrt{5}\right)A{e}^{\left(3+\sqrt{5}\right)x}+\left(3-\sqrt{5}\right)B{e}^{\left(3-\sqrt{5}\right)x}$
so we have $3\left(A+B\right)+\sqrt{5}\left(A-B\right)=18$ which gives $A=B=3$ and $y=3{e}^{\left(3+\sqrt{5}\right)x}+3{e}^{\left(3-\sqrt{5}\right)x}\phantom{\rule{0.3em}{0ex}}.$
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is incorrect
Find the general solution to the second order linear differential equation
$\frac{{d}^{2}y}{d{x}^{2}}-6\frac{dy}{dx}+13y=0.$
 a) $y={e}^{2x}\left(Asin3x+Bcos3x\right)$ b) $y={e}^{3x}\left(Asin2x+Bcos2x\right)$ c) $y={e}^{-2x}\left(Asin3x+Bcos3x\right)$ d) $y=A{e}^{x}+B{e}^{5x}$

Choice (a) is incorrect
Choice (b) is correct!
${k}^{2}-6k+13=0$ has solutions $k=3±2\sqrt{-1}$ hence $y={e}^{3x}\left(Asin2x+Bcos2x\right)\phantom{\rule{0.3em}{0ex}}.$
Choice (c) is incorrect
Choice (d) is incorrect
Find the particular solution of the second order linear differential equation
$\frac{{d}^{2}y}{d{t}^{2}}-8\frac{dy}{dt}+25y=0$
where $y=-3$ and $\frac{dy}{dt}=-6$ when $t=0\phantom{\rule{0.3em}{0ex}}.$
 a) $y={e}^{4t}\left(2sin3t-3cos3t\right)$ b) $y={e}^{3t}\left(2sin4t-3cos4t\right)$ c) $y=-{e}^{-4t}\left(2sin3t+3cos3t\right)$ d) $y=-{e}^{-3t}\left(2sin4t+3cos4t\right)$

Choice (a) is correct!
${k}^{2}-8k+25=0$ has solutions $k=4±3\sqrt{-1}$ hence $y={e}^{4t}\left(Asin3t+Bcos3t\right)\phantom{\rule{0.3em}{0ex}}.$ At $t=0$ we have $y=B=-3$ and
$\frac{dy}{dt}=4{e}^{4t}\left(Asin3t+Bcos3t\right)+{e}^{4t}\left(3Acos3t-3Bsin3t\right)$
hence $-12+3A=-6$ and $A=2$. The particular solution is thus
$y={e}^{4t}\left(2sin3t-3cos3t\right)\phantom{\rule{0.3em}{0ex}}.$
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is incorrect
Find the general solution to the second order linear differential equation
$\frac{{d}^{2}y}{d{t}^{2}}-25y=0\phantom{\rule{0.3em}{0ex}}.$
 a) $y=Acos5t+Bsin5t$ b) $y=A{e}^{-5t}+B{e}^{5t}$ c) $y=5cost+5sint$ d) $y=A+B{e}^{25t}$

Choice (a) is incorrect
Choice (b) is correct!
${k}^{2}-25=\left(k-5\right)\left(k+5\right)=0$ hence $k=5$ or $k=-5$ and the general solution is $y=A{e}^{-5t}+B{e}^{5t}\phantom{\rule{0.3em}{0ex}}.$
Choice (c) is incorrect
Choice (d) is incorrect
Find the particular solution to the second order linear differential equation
$\frac{{d}^{2}x}{d{t}^{2}}-6\frac{dx}{dt}+8x=0,$
where $x=-3$ and $\frac{dx}{dt}=-6$ when $t=0\phantom{\rule{0.3em}{0ex}}.$
 a) $x=A{e}^{2t}+B{e}^{4t}$ b) $x=A{e}^{-2t}+B{e}^{-4t}$ c) $x=-\frac{3}{2}{e}^{2t}-\frac{3}{2}{e}^{4t}$ d) $x=-3{e}^{2t}$

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is correct!
${k}^{2}-6k+8=0$ has solutions $k=2\phantom{\rule{0.3em}{0ex}}$, $k=4$ hence the general solution is $x=A{e}^{2t}+B{e}^{4t}\phantom{\rule{0.3em}{0ex}}.$ Since $x=-3$ when $t=0$ we have $A+B=-3$
$\frac{dx}{dt}=2A{e}^{2t}+4B{e}^{4t}$
so we have $2A+4B=-6$ which gives $A=-3$ and $B=0\phantom{\rule{0.3em}{0ex}}.$ Hence $x=-3{e}^{2t}\phantom{\rule{0.3em}{0ex}}.$