Q107 / 180 · COMEDK 2024 (Morning Shift) · Mathematics
The particular solution of the differential equation $\cos x \dfrac{d y}{d x}+y=\sin x$ at $y(0)=1$
- A$y(\sec x+\tan x)=\sec x+\tan x-x+1$
- B$y(\sec x+\tan x)=\sec x+\tan x-x+2$
- Correct answerC$y(\sec x+\tan x)=\sec x+\tan x-x$
- D$y(\sec x+\tan x)=\sec x+\tan x+x$
Solution
The given differential equation is $\cos x \dfrac{dy}{dx} + y = \sin x$. Dividing by $\cos x$, we get $\dfrac{dy}{dx} + y \sec x = \tan x$.
This is a linear differential equation of the form $\dfrac{dy}{dx} + Py = Q$, where $P = \sec x$ and $Q = \tan x$.
The integrating factor (IF) is $e^{\int P dx} = e^{\int \sec x dx} = e^{\ln|\sec x + \tan x|} = \sec x + \tan x$.
The general solution is $y \cdot (IF) = \int Q \cdot (IF) dx + C$.
$y(\sec x + \tan x) = \int \tan x (\sec x + \tan x) dx + C$.
$y(\sec x + \tan x) = \int (\sec x \tan x + \tan^2 x) dx + C$.
Using $\tan^2 x = \sec^2 x - 1$, we have $y(\sec x + \tan x) = \int (\sec x \tan x + \sec^2 x - 1) dx + C$.
$y(\sec x + \tan x) = \sec x + \tan x - x + C$.
Given $y(0) = 1$, substituting $x = 0$ and $y = 1$: $1(\sec 0 + \tan 0) = \sec 0 + \tan 0 - 0 + C$.
$1(1 + 0) = 1 + 0 - 0 + C$, which implies $1 = 1 + C$, so $C = 0$.
Therefore, the particular solution is $y(\sec x + \tan x) = \sec x + \tan x - x$.
Answer: $y(\sec x+\tan x)=\sec x+\tan x-x$
This is a linear differential equation of the form $\dfrac{dy}{dx} + Py = Q$, where $P = \sec x$ and $Q = \tan x$.
The integrating factor (IF) is $e^{\int P dx} = e^{\int \sec x dx} = e^{\ln|\sec x + \tan x|} = \sec x + \tan x$.
The general solution is $y \cdot (IF) = \int Q \cdot (IF) dx + C$.
$y(\sec x + \tan x) = \int \tan x (\sec x + \tan x) dx + C$.
$y(\sec x + \tan x) = \int (\sec x \tan x + \tan^2 x) dx + C$.
Using $\tan^2 x = \sec^2 x - 1$, we have $y(\sec x + \tan x) = \int (\sec x \tan x + \sec^2 x - 1) dx + C$.
$y(\sec x + \tan x) = \sec x + \tan x - x + C$.
Given $y(0) = 1$, substituting $x = 0$ and $y = 1$: $1(\sec 0 + \tan 0) = \sec 0 + \tan 0 - 0 + C$.
$1(1 + 0) = 1 + 0 - 0 + C$, which implies $1 = 1 + C$, so $C = 0$.
Therefore, the particular solution is $y(\sec x + \tan x) = \sec x + \tan x - x$.
Answer: $y(\sec x+\tan x)=\sec x+\tan x-x$
Hence, the correct answer is option: 'C'.