JEE Main 2024 (Online) 5th April Morning ShiftQuestion 1
If $y=y(x)$ is the solution of the differential equation $\dfrac{dy}{dx}+2y=\sin(2x)$, $y(0)=\dfrac{3}{4}$, then $y\left(\dfrac{\pi}{8}\right)$ is equal to :
Differential Equation
Solution
Concept. Solution $=$ $Ce^{-2x}$ plus a particular solution $A\sin2x+B\cos2x$. Formulas used. (1) $y'+2y=g(x)$: $y=Ce^{-2x}+y_p$Step 1 - Particular solution. Put $y_p=A\sin2x+B\cos2x$: $$(2A-2B)\sin2x+(2A+2B)\cos2x=\sin2x\ \Longrightarrow\ A=\tfrac14,\ B=-\tfrac14.$$ Step 2 - General solution. By (1), $y=Ce^{-2x}+\dfrac{\sin2x-\cos2x}{4}$. Step 3 - Constant. $y(0)=C-\frac14=\frac34$, so $C=1$. Step 4 - At $x=\frac\pi8$. $\sin\frac\pi4=\cos\frac\pi4$, so $y=e^{-\pi/4}$. Answer: (A)