Differential Equations 2 Question 10
10. If $y(x)$ is the solution of the differential equation
$$ \frac{d y}{d x}+\frac{2 x+1}{x} y=e^{-2 x}, x>0 $$
where $y(1)=\frac{1}{2} e^{-2}$, then
(2019 Main, 11 Jan I)
(a) $y(x)$ is decreasing in $\frac{1}{2}, 1$
(b) $y(x)$ is decreasing in $(0,1)$
(c) $y\left(\log _e 2\right)=\log _e 4$
(d) $y\left(\log _e 2\right)=\frac{\log _e 2}{4}$
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Answer:
Correct Answer: 10. (a)
Solution:
- We have, $\frac{d y}{d x}+\frac{2 x+1}{x} y=e^{-2 x}$
which is of the form $\frac{d y}{d x}+P y=Q$, where
$$ P=\frac{2 x+1}{x} \text { and } Q=e^{-2 x} $$
Now, IF $=e^{\int P d x}=e^{\int \frac{1+2 x}{x} d x}=e^{\int \frac{1}{x}+2 d x}$
$$ =e^{\ln x+2 x}=e^{\ln x} \cdot e^{2 x}=x \cdot e^{2 x} $$
and the solution of the given equation is
$$ \begin{aligned} y \cdot(IF) & =\int(IF) Q d x+C \\ \Rightarrow \quad y\left(x e^{2 x}\right) & =\int\left(x e^{2 x} \cdot e^{-2 x}\right) d x+C \\ & =\int x d x+C=\frac{x^{2}}{2}+C \end{aligned} $$
Since, $y=\frac{1}{2} e^{-2}$ when $x=1$
$\therefore \quad \frac{1}{2} e^{-2} \cdot e^{2}=\frac{1}{2}+C \Rightarrow C=0$ (using Eq. (i))
$\therefore \quad y\left(x e^{2 x}\right)=\frac{x^{2}}{2} \quad \Rightarrow \quad y=\frac{x}{2} e^{-2 x}$
Now, $\frac{d y}{d x}=\frac{1}{2} e^{-2 x}+\frac{x}{2} e^{-2 x}(-2)=e^{-2 x} \frac{1}{2}-x \quad<0$,
if $\frac{1}{2}<x<1$
[by using product rule of derivative]
and $y\left(\log _e 2\right)=\frac{\log _e 2}{2} e^{-2 \log _e 2}=\frac{1}{2} \log _e 2 e^{\log _e 2^{-2}}$
$$ =\frac{1}{2} \cdot \log _e 2 \cdot 2^{-2}=\frac{1}{8} \log _e 2 $$
