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JEE Main · MCQ · Differential Equations
MCQ · Mathematics · Differential Equations
Q. Let $y = y(x)$ be the solution of the differential equation
$$\sec x\,\dfrac{dy}{dx} – 2y = 2 + 3\sin x,\quad x \in \left(-\dfrac{\pi}{2},\,\dfrac{\pi}{2}\right)$$
$y(0) = -\dfrac{7}{4}$. Then $y\!\left(\dfrac{\pi}{6}\right)$ is equal to :
A$-\dfrac{5}{2}$ ✓
B$-3\sqrt{2} – 7$
C$-\dfrac{5}{4}$
D$-3\sqrt{3} – 7$
✅ Correct Answer: (A) $-\dfrac{5}{2}$
Step-by-Step Solution
1
Convert to standard linear form
Multiply through by $\cos x$:
$\dfrac{dy}{dx} – 2\cos x\cdot y = (2 + 3\sin x)\cos x$
Here $P = -2\cos x$, $Q = (2+3\sin x)\cos x$
2
Integrating Factor
$\text{IF} = e^{\int -2\cos x\,dx} = e^{-2\sin x}$
3
Multiply and integrate
$y\cdot e^{-2\sin x} = \displaystyle\int (2+3\sin x)\cos x\cdot e^{-2\sin x}\,dx$
Substitute $t = \sin x$, $dt = \cos x\,dx$:
$= \displaystyle\int (2+3t)e^{-2t}\,dt$
4
Evaluate $\int(2+3t)e^{-2t}\,dt$ by IBP
$= (2+3t)\cdot\dfrac{e^{-2t}}{-2} – \int 3\cdot\dfrac{e^{-2t}}{-2}\,dt$
$= -\dfrac{(2+3t)}{2}e^{-2t} – \dfrac{3}{4}e^{-2t} + C$
$= -e^{-2t}\!\left(\dfrac{2+3t}{2} + \dfrac{3}{4}\right) + C$
$= -e^{-2t}\cdot\dfrac{4+6t+3}{4} + C = -\dfrac{(6t+7)}{4}e^{-2t} + C$
$= -\dfrac{(2+3t)}{2}e^{-2t} – \dfrac{3}{4}e^{-2t} + C$
$= -e^{-2t}\!\left(\dfrac{2+3t}{2} + \dfrac{3}{4}\right) + C$
$= -e^{-2t}\cdot\dfrac{4+6t+3}{4} + C = -\dfrac{(6t+7)}{4}e^{-2t} + C$
5
General solution (back-substitute $t = \sin x$)
$y\cdot e^{-2\sin x} = -\dfrac{(6\sin x+7)}{4}e^{-2\sin x} + C$
$\Rightarrow y = -\dfrac{6\sin x + 7}{4} + C\,e^{2\sin x}$
$\Rightarrow y = -\dfrac{6\sin x + 7}{4} + C\,e^{2\sin x}$
6
Apply $y(0) = -\dfrac{7}{4}$
$-\dfrac{7}{4} = -\dfrac{7}{4} + C\cdot e^0$
$\Rightarrow C = 0$
$\Rightarrow C = 0$
7
Particular solution and find $y(\pi/6)$
$y = -\dfrac{6\sin x + 7}{4}$
At $x = \pi/6$: $\sin(\pi/6) = 1/2$:
$y\!\left(\dfrac{\pi}{6}\right) = -\dfrac{6\cdot\frac{1}{2}+7}{4} = -\dfrac{3+7}{4} = -\dfrac{10}{4} = -\dfrac{5}{2}$
$y(\pi/6) = -5/2$ → Option (A) ✓
Related Theory
📌 Linear ODE with Trigonometric IF
When sec x appears as coefficient of dy/dx, multiply by cos x to convert to standard form. IF = e^(∫P dx). Here P = −2cosx → IF = e^(−2sinx).
IF = e^(−2sinx)∫−2cosx dx = −2sinx
IF = e^(−2sinx)∫−2cosx dx = −2sinx
📌 IBP Result
∫(2+3t)e^(−2t)dt = −(6t+7)e^(−2t)/4 + C
📌 Why C = 0
The initial condition y(0) = −7/4 exactly matches the particular solution at x=0, forcing C=0. This is an elegant result — the particular solution alone satisfies the IVP.
Frequently Asked Questions
1. How is the ODE converted to standard form?
Multiply by cosx: dy/dx − 2cosx·y = (2+3sinx)cosx.
2. What is the IF?
e^(−2sinx).
3. What substitution is used?
t = sinx, dt = cosx dx.
4. What is the general solution?
y = −(6sinx+7)/4 + C·e^(2sinx).
5. Why is C = 0?
y(0) = −7/4 = −7/4 + C → C = 0.
6. What is sin(π/6)?
sin(π/6) = 1/2.
7. What is y(π/6)?
y(π/6) = −(3+7)/4 = −10/4 = −5/2.
8. What is the IBP formula used?
∫u dv = uv − ∫v du. Here u = (2+3t), dv = e^(−2t)dt.
9. What is the correct answer?
Option (A): −5/2.
10. What makes this ODE non-trivial?
The sec x coefficient and combined (2+3sinx)cosx RHS requiring substitution t=sinx after IBP.
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