Question

Let 𝑦: ℝ → ℝ be the solution to the differential equation

$\frac{d^{2}y}{d{x}^2}+2⁤\frac{dy}{dx}+5y=1$

satisfying 𝑦(0) = 0 and 𝑦′(0) = 1.
Then, $\underset{x\to \infty }{lim}y\left(x\right)$ equals _____________ (rounded off to two decimal places).

Answer: $\underset{\_\_\_\_\_\_\_\_\_\_\_\_\_}{0.20}$

Solution :-

Given that $\frac{d^{2}y}{d{x}^2}+2⁤\frac{dy}{dx}+5y=1$ this is a 2nd order linear differential equation with constant cofficient and we know that general solution of heigher order linear differential equation is $y=C\mathrm{.}F\mathrm{.}\left(\text{complementary function}\right)+P\mathrm{.}I\mathrm{.}\left(\text{particular integral}\right)\mathrm{.}$

We can write $\frac{d^{2}y}{d{x}^2}+2⁤\frac{dy}{dx}+5y=1$ as $\frac{d^{2}y}{d{x}^2}+2⁤\frac{dy}{dx}+5y=e^{0\cdot x}\mathrm{.}$

Auxiliary Equation of given differential equation is

Roots of Auxiliary equation is complex.
Hence, $C\mathrm{.}F\mathrm{.}=e^{-x}\left(c_{1}cos2x+c_{2}sin2x\right)$

Now, $P\mathrm{.}I\mathrm{.}=\frac{1}{F\left(D\right)}\cdot Q\left(x\right)$ here $F\left(D\right)=D^2+2D+5$ and $Q\left(x\right)=e^{0\cdot x}\mathrm{.}$

So, the general solution is $y\left(x\right)=C\mathrm{.}F\mathrm{.}+P\mathrm{.}I\mathrm{.}=e^{-x}\left(c_{1}cos2x+c_{2}sin2x\right)+\frac{1}{5}\mathrm{.}$

Now, given that $y\left(0\right)=0$ , hence from the general solution we get

Now, given that $y^{\prime }\left(0\right)=1$ , hence from the general solution we get

Now, put the value of $c_1$ and $c_2$ in the general solution we get final solution as

So, if $x\to \infty$ then $-\frac{1}{5}cos2x+\frac{2}{5}sin2x\in \left(-\frac{3}{5},\frac{3}{5}\right)$ but $e^x\to \infty \mathrm{.}$

So, the correct answer is $\to$ $\underset{\_\_\_\_\_\_\_\_\_\_\_\_\_}{0.20}$