Differential Equations
Differential Equations
nta_pyq_2025_apr
Grade 12
Question:
If for the solution curve $y = f(x)$ of the differential equation $\dfrac{dy}{dx}+(\tan x)y = \dfrac{2+\sec x}{(1+2\sec x)^2}$, $x\in\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$, $f\!\left(\dfrac{\pi}{3}\right) = \dfrac{\sqrt{3}}{10}$, then $f\!\left(\dfrac{\pi}{4}\right)$ is equal to:
$\dfrac{\sqrt{3}+1}{10(4+\sqrt{3})}$
$\dfrac{5-\sqrt{3}}{2\sqrt{2}}$
$\dfrac{9\sqrt{3}+3}{10(4+\sqrt{3})}$
$\dfrac{4-\sqrt{2}}{14}$
Step-by-Step Solution
Key Concept: I.F. $= e^{\int\tan x\,dx} = \sec x$; multiply through and integrate $\int\sec x\cdot\dfrac{2+\sec x}{(1+2\sec x)^2}dx$ using the substitution $\cos x = \tfrac{1-t^2}{1+t^2}$ (Weierstrass).
I.F. $= \sec x$. Multiply: $\dfrac{d}{dx}(y\sec x) = \dfrac{(2+\sec x)\sec x}{(1+2\sec x)^2}$.
Let $t = \tan(x/2)$, $\cos x = \tfrac{1-t^2}{1+t^2}$, $\sec x = \tfrac{1+t^2}{1-t^2}$, $dx = \tfrac{2dt}{1+t^2}$.
After substitution and simplification: $y\sec x = \dfrac{2}{t+3/t}+C$.
At $x=\pi/3$, $t=\tan(\pi/6)=1/\sqrt{3}$, $y=\sqrt{3}/10$: $C=0$.
At $x=\pi/4$, $t=\sqrt{2}-1$:
$$y\cdot\sqrt{2} = \frac{2}{\sqrt{2}-1+\frac{3}{\sqrt{2}-1}} = \frac{2(\sqrt{2}-1)}{(\sqrt{2}-1)^2+3} = \frac{2(\sqrt{2}-1)}{6-2\sqrt{2}} = \frac{\sqrt{2}-1}{3-\sqrt{2}}.$$
$$y = \frac{\sqrt{2}-1}{\sqrt{2}(3-\sqrt{2})} = \frac{2\sqrt{2}-2}{2(6-2\sqrt{2})} = \frac{4-\sqrt{2}}{14}.$$
Correct Answer: 4