Question

Solve for $y$ if $\frac{d^{2}y}{d{t}^2}+2\frac{dy}{dt}+y=0$ with $y\left(0\right)=1$ and $y^{\prime }\left(0\right)=-2$

• A. $(1-t)e^{-t}$
• B. $(1+t)e^t$
• C. $(1+t)e^{-t}$
• D. $(1-t)e^t$

Answer 1

The correct option is A $(1-t)e^{-t}$

Given $(D^2+2D+1)y=0$
So, AE is
$m^2+2m+1=0$
$(m+1{)}^2=0$
$m=-1,-1$ (real and equal)
$\therefore$ General solution is
$y=(C_1+C_{2}t)e^{-t}$ .... (i)
$\frac{dy}{dt}=(C_1+C_{2}t)(-e^{-t})+C_2{e}^{-t}$ ...(ii)
Now using $y\left(0\right)=1$ in (ii), $C_1=1$
Again using $y^{\prime }\left(0\right)=-2$ in (iii), $C_2=-1$
Hence the solution is $y=(1-t)e^{-t}$