Differential Equations Ques 43
Let $y=y(x)$ be the solution of the differential equation, $\frac{d y}{d x}+y \tan x=2 x+x^{2} \tan x, \quad x \in(-\frac{\pi}{2}, \frac{\pi}{2})$, such that $y(0)=1$. Then
(2019 Main, 10 April II)
(a) $y^{\prime} (\frac{\pi}{4})-y^{\prime}(-\frac{\pi}{4})=\pi-\sqrt{2}$
(b) $y^{\prime} (\frac{\pi}{4})+y^{\prime}(-\frac{\pi}{4})=-\sqrt{2}$
(c) $y (\frac{\pi}{4})+y(-\frac{\pi}{4})=\frac{\pi^{2}}{2}+2$
(d) $y (\frac{\pi}{4})-y(-\frac{\pi}{4})=\sqrt{2}$
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Answer:
Correct Answer: 43.(a)
Solution:
Formula:
ELEMENTARY TYPES OF FIRST ORDER \& FIRST DEGREE DIFFERENTIAL EQUATIONS :
- Given differential equation is
$\frac{d y}{d x}+y \tan x=2 x+x^{2} \tan x$, which is linear differential equation in the form of $\frac{d y}{d x}+P y=Q$.
Here, $P=\tan x$ and $Q=2 x+x^{2} \tan x$
$\therefore IF=e^{\int \tan x d x}=e^{\log _e(\sec x)}=\sec x$
Now, solution of linear differential equation is given as
$ \begin{aligned} & y \times IF=\int(Q \times IF) d x+C \\ & \therefore y(\sec x)=\int\left(2 x+x^{2} \tan x\right) \sec x d x+C \\ & =\int(2 x \sec x) d x+\int x^{2} \sec x \tan x d x+C \\ & \because \int x^{2} \sec x \tan x d x=x^{2} \sec x-\int(2 x \sec x) d x \end{aligned} $
Therefore, solution is
$y \sec x=2 \int x \sec x d x+x^{2} \sec x-2 \int x \sec x d x+C$
$\Rightarrow y \sec x=x^{2} \sec x+C \ldots$ (i)
$\because y(0)=1 \Rightarrow 1(1)=0(1)+C \Rightarrow C=1$
Now, $y=x^{2}+\cos x$
and $y^{\prime}=2 x-\sin x$
According to options,
$y^{\prime} (\frac{\pi}{4})-y^{\prime} (\frac{-\pi}{4})=(2 \frac{\pi}{4}-\frac{1}{\sqrt{2}})$
$-(2(-\frac{\pi}{4})+\frac{1}{\sqrt{2}})=\pi-\sqrt{2}$
and $y^{\prime} (\frac{\pi}{4})+y^{\prime}(-\frac{\pi}{4})=2 \frac{\pi}{4}-\frac{1}{\sqrt{2}}+2-\frac{\pi}{4}+\frac{1}{\sqrt{2}}=0$
and $y (\frac{\pi}{4})+y(-\frac{\pi}{4})=\frac{\pi^{2}}{16}+\frac{1}{\sqrt{2}}+\frac{\pi^{2}}{16}+\frac{1}{\sqrt{2}}=\frac{\pi^{2}}{4}+\sqrt{2}$
and $y (\frac{\pi}{4})-y(-\frac{\pi}{4})=\frac{\pi^{2}}{16}+\frac{1}{\sqrt{2}}-\frac{\pi^{2}}{16}-\frac{1}{\sqrt{2}}=0$
BETA
