Question

For initial value problem $\stackrel{..}{y}+2\stackrel{.}{y}++\left(101\right)y=\left(10.4\right)e^x,y\left(0\right)=1.1$ and $\stackrel{.}{y}\left(0\right)=-0.9$. Various solutions are writen in the following groups. Match the type of solution with the correct expression.
Group - 1
P. General solution of homogeneous equations
Q. Particular integral
R. Total solution satisfying boundary conditions.
Group - II
(1) $0.1e^x$
(2) $e^{-x}[Acos10x+Bsin10x]$
(3) $e^{-x}cos10x+0.1e^x$

• A. $P-2,Q-1,R-3$
• B. $P-1,Q-3,R-2$
• C. $P-1,Q-2,R-3$
• D. $P-3,Q-2,R-1$
  

Answer 1

The correct option is A $P-2,Q-1,R-3$

Given $(D^2+2D+101)y=10.4e^x$ ... (i)
So $AE$ is $m^2+2m+101=0$
$\Rightarrow m=-1\pm 10i$
$C.F.$ is $e^{-x}[c_{1}cos10x+c_{2}sin10x]$
$P.I.=\frac{1}{(D^2+2D+101)}\left(10.4\right)e^x$
$=\frac{1}{(1+2+101)}\left(10.4\right)e^x=0.1e^x$
So general solution is
$y=CF+PI$
$y=e^{-x}\left[c_{1}cos\right(10)x+c_{2}sin(10\left)x\right]+\left(0.1\right)e^x$ ... (ii)
$\frac{dy}{dx}=-e^{-x}[c_{1}cos10x+c_{2}sin10x]+0.1e^x+e^{-x}[-c_{1}10cos10x+c_{2}10cos10x]$
... (iii)
Now using $y\left(0\right)=1.1$ & $y^{\prime }\left(0\right)=-0.9$ in (2) & (3), we get $C_1=1$ & $C_2=0$
So, solution is
$y=e^{-x}\left[cos\right(10x\left)\right]+\left(0.1\right)e^x$
Hence, $P-2,Q-1,R-3$ is the correct option.