If for the solution curve y=f(x) of the differential equation \fracd yd x+(\tan x) y=\frac2+\sec x(1+2 \sec x)^2 , x \in(\frac-\pi2, \frac\pi2), f(\frac\pi3)=\frac\sqrt310 , then f(\frac\pi4) is equal to:

Differential Equations PYQ — JEE Main 2025 · 29 Jan · Shift 2 · Q38 with Solution

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Correct answer: B

Given differential equation

$\frac{dy}{dx}+ (\tan x) y=\frac{2+\sec x}{(1+2\sec x)^2}, \qquad x\in\left(-\frac\pi2,\frac\pi2\right)$

We must find $f\left(\frac\pi4\right)$ given

$f\left(\frac\pi3\right)=\frac{\sqrt3}{10}.$

Solve the linear differential equation

This is a first-order linear ODE:

$\frac{dy}{dx}+P(x)y=Q(x), \quad P(x)=\tan x.$

The integrating factor is

$\text{I.F.}=e^{\int \tan x\,dx}=e^{-\ln(\cos x)}=\sec x$

since on $\left(-\frac\pi2,\frac\pi2\right)$ , $\cos x>0$ .

Multiplying the equation by $\sec x$ :

$\sec x\frac{dy}{dx}+\sec x\tan x\,y=\frac{(2+\sec x)\sec x}{(1+2\sec x)^2}.$

The left side is

$\frac{d}{dx}(y\sec x).$

Hence,

$\frac{d}{dx}(y\sec x)=\frac{(2+\sec x)\sec x}{(1+2\sec x)^2}.$

Integrate the right-hand side

Let

$t=1+2\sec x.$

Then

$dt=2\sec x\tan x\,dx,$

which is not directly helpful. So instead, observe a better approach:

Try to identify the RHS as a derivative involving $\tan x$ and $1+2\sec x$ .

Consider

$\frac{d}{dx}\left(\frac{\sin x}{1+2\sec x}\right).$

Now,

$\frac{\sin x}{1+2\sec x}=\frac{\sin x\cos x}{\cos x+2}.$

Differentiate directly:

$\frac{d}{dx}(y\sec x)=\frac{(2+\sec x)\sec x}{(1+2\sec x)^2}.$

Let us rewrite the RHS in terms of $\cos x$ :

$\sec x=\frac1{\cos x}.$

Then

=\frac{\left(2+\frac1{\cos x}\right)\frac1{\cos x}}{\left(1+\frac2{\cos x}\right)^2}.

Multiply numerator and denominator by $\cos^2 x$ :

$=\frac{(2\cos x+1)}{(\cos x+2)^2}.$

So we need to integrate

$\frac{d}{dx}(y\sec x)=\frac{2\cos x+1}{(\cos x+2)^2}.$

Now check

=\frac{\cos x(2+\cos x)-\sin x(-\sin x)}{(2+\cos x)^2} =\frac{2\cos x+\cos^2 x+\sin^2 x}{(2+\cos x)^2} =\frac{2\cos x+1}{(2+\cos x)^2}.

This matches exactly.

Therefore,

$\frac{d}{dx}(y\sec x)=\frac{d}{dx}\left(\frac{\sin x}{2+\cos x}\right).$

Integrating,

$y\sec x=\frac{\sin x}{2+\cos x}+C.$

Thus,

$y=\cos x\left(\frac{\sin x}{2+\cos x}+C\right).$

Use the condition $f\left(\frac\pi3\right)=\frac{\sqrt3}{10}$

At $x=\frac\pi3$ ,

$\sin\frac\pi3=\frac{\sqrt3}{2}, \qquad \cos\frac\pi3=\frac12, \qquad \sec\frac\pi3=2.$

Using

$y\sec x=\frac{\sin x}{2+\cos x}+C,$

we get

$\frac{\sqrt3}{10}\cdot 2=\frac{\frac{\sqrt3}{2}}{2+\frac12}+C.$

So,

=\frac{\sqrt3}{5}+C.$$ Hence, $$C=0.$$ Therefore the solution curve is $$y=\cos x\cdot \frac{\sin x}{2+\cos x} =\frac{\sin x\cos x}{2+\cos x}.$$ --- 5. **Find $f\left(\frac\pi4\right)$** At $x=\frac\pi4$, $$\sin\frac\pi4=\cos\frac\pi4=\frac1{\sqrt2}.$$ Therefore, $$f\left(\frac\pi4\right)=\frac{\frac1{\sqrt2}\cdot\frac1{\sqrt2}}{2+\frac1{\sqrt2}} =\frac{\frac12}{2+\frac1{\sqrt2}}.$$ Now simplify: $$\frac{1}{2\left(2+\frac1{\sqrt2}\right)} =\frac{1}{4+\sqrt2}.$$ Rationalizing, $$\frac{1}{4+\sqrt2}=\frac{4-\sqrt2}{16-2}=\frac{4-\sqrt2}{14}.$$ So, $$f\left(\frac\pi4\right)=\frac{4-\sqrt2}{14}.$$ --- 6. **Compare with options** This matches **Option B**: $$\boxed{\frac{4-\sqrt2}{14}}$$ --- 7. **Comparison with stored answer** Stored correct answer: **B** Our derived answer is also **B**, so they agree.

More from Differential Equations

View written solutionFree

Correct answer: B

Given differential equation

$\frac{dy}{dx}+ (\tan x) y=\frac{2+\sec x}{(1+2\sec x)^2}, \qquad x\in\left(-\frac\pi2,\frac\pi2\right)$

We must find $f\left(\frac\pi4\right)$ given

$f\left(\frac\pi3\right)=\frac{\sqrt3}{10}.$

Solve the linear differential equation

This is a first-order linear ODE:

$\frac{dy}{dx}+P(x)y=Q(x), \quad P(x)=\tan x.$

The integrating factor is

$\text{I.F.}=e^{\int \tan x\,dx}=e^{-\ln(\cos x)}=\sec x$

since on $\left(-\frac\pi2,\frac\pi2\right)$ , $\cos x>0$ .

Multiplying the equation by $\sec x$ :

$\sec x\frac{dy}{dx}+\sec x\tan x\,y=\frac{(2+\sec x)\sec x}{(1+2\sec x)^2}.$

The left side is

$\frac{d}{dx}(y\sec x).$

Hence,

$\frac{d}{dx}(y\sec x)=\frac{(2+\sec x)\sec x}{(1+2\sec x)^2}.$

Integrate the right-hand side

Let

$t=1+2\sec x.$

Then

$dt=2\sec x\tan x\,dx,$

which is not directly helpful. So instead, observe a better approach:

Try to identify the RHS as a derivative involving $\tan x$ and $1+2\sec x$ .

Consider

$\frac{d}{dx}\left(\frac{\sin x}{1+2\sec x}\right).$

Now,

$\frac{\sin x}{1+2\sec x}=\frac{\sin x\cos x}{\cos x+2}.$

Differentiate directly:

$\frac{d}{dx}(y\sec x)=\frac{(2+\sec x)\sec x}{(1+2\sec x)^2}.$

Let us rewrite the RHS in terms of $\cos x$ :

$\sec x=\frac1{\cos x}.$

Then

=\frac{\left(2+\frac1{\cos x}\right)\frac1{\cos x}}{\left(1+\frac2{\cos x}\right)^2}.

Multiply numerator and denominator by $\cos^2 x$ :

$=\frac{(2\cos x+1)}{(\cos x+2)^2}.$

So we need to integrate

$\frac{d}{dx}(y\sec x)=\frac{2\cos x+1}{(\cos x+2)^2}.$

Now check

=\frac{\cos x(2+\cos x)-\sin x(-\sin x)}{(2+\cos x)^2} =\frac{2\cos x+\cos^2 x+\sin^2 x}{(2+\cos x)^2} =\frac{2\cos x+1}{(2+\cos x)^2}.

This matches exactly.

Therefore,

$\frac{d}{dx}(y\sec x)=\frac{d}{dx}\left(\frac{\sin x}{2+\cos x}\right).$

Integrating,

$y\sec x=\frac{\sin x}{2+\cos x}+C.$

Thus,

$y=\cos x\left(\frac{\sin x}{2+\cos x}+C\right).$

Use the condition $f\left(\frac\pi3\right)=\frac{\sqrt3}{10}$

At $x=\frac\pi3$ ,

$\sin\frac\pi3=\frac{\sqrt3}{2}, \qquad \cos\frac\pi3=\frac12, \qquad \sec\frac\pi3=2.$

Using

$y\sec x=\frac{\sin x}{2+\cos x}+C,$

we get

$\frac{\sqrt3}{10}\cdot 2=\frac{\frac{\sqrt3}{2}}{2+\frac12}+C.$

So,

=\frac{\sqrt3}{5}+C.$$ Hence, $$C=0.$$ Therefore the solution curve is $$y=\cos x\cdot \frac{\sin x}{2+\cos x} =\frac{\sin x\cos x}{2+\cos x}.$$ --- 5. **Find $f\left(\frac\pi4\right)$** At $x=\frac\pi4$, $$\sin\frac\pi4=\cos\frac\pi4=\frac1{\sqrt2}.$$ Therefore, $$f\left(\frac\pi4\right)=\frac{\frac1{\sqrt2}\cdot\frac1{\sqrt2}}{2+\frac1{\sqrt2}} =\frac{\frac12}{2+\frac1{\sqrt2}}.$$ Now simplify: $$\frac{1}{2\left(2+\frac1{\sqrt2}\right)} =\frac{1}{4+\sqrt2}.$$ Rationalizing, $$\frac{1}{4+\sqrt2}=\frac{4-\sqrt2}{16-2}=\frac{4-\sqrt2}{14}.$$ So, $$f\left(\frac\pi4\right)=\frac{4-\sqrt2}{14}.$$ --- 6. **Compare with options** This matches **Option B**: $$\boxed{\frac{4-\sqrt2}{14}}$$ --- 7. **Comparison with stored answer** Stored correct answer: **B** Our derived answer is also **B**, so they agree.

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