JEE Main 2025 (January)

LEVELJEE Main

Animated Solution for Mathematics - Differential Equations: Let $y = y (x)$ be the solution of the differential equation $cos x (lo g_{e} (cos x))^{2} d y + (sin x - 3 y sin x lo g_{e} (cos x)) d x = 0$ $x \in (0, \frac{π}{2})$ . If $y (\frac{π}{4}) = \frac{- 1}{l o g _{e} 2}$ , then $y (\frac{π}{6})$ is equal to :

Q: How did you know to substitute $t = \ln(\cos x)$? I never see these substitutions.

It comes from practice! Look at the coefficient of $y$, which is $P(x) = -\frac{3\tan x}{\ln(\cos x)}$. The numerator is $-\tan x$, which is exactly the derivative of $\ln(\cos x)$. Whenever you see a function and its derivative in the same expression, that is a huge signal to use substitution.

Select Answer:

Visualized Solution

Analyze the Differential Equation

Given equation: $cos x (lo g_{e} (cos x))^{2} d y + (sin x - 3 y sin x lo g_{e} (cos x)) d x = 0$
Goal: Transform into the standard form $\frac{d y}{d x} + P (x) y = Q (x)$

Rearranging to Standard Form

Divide by $d x$ : $cos x (lo g_{e} (cos x))^{2} \frac{d y}{d x} + sin x - 3 y sin x lo g_{e} (cos x) = 0$
Rearrange terms: $cos x (lo g_{e} (cos x))^{2} \frac{d y}{d x} - 3 y sin x lo g_{e} (cos x) = - sin x$

Simplifying Coefficients

Divide by $cos x (lo g_{e} (cos x))^{2}$ :
$\frac{d y}{d x} - \frac{3 s i n x l o g _{e} ( c o s x )}{c o s x ( l o g _{e} ( c o s x ) ) ^{2}} y = \frac{- s i n x}{c o s x ( l o g _{e} ( c o s x ) ) ^{2}}$
Simplified form: $\frac{d y}{d x} - \frac{3 t a n x}{l o g _{e} ( c o s x )} y = \frac{- t a n x}{( l o g _{e} ( c o s x ) ) ^{2}}$

Identifying $P (x)$ and $Q (x)$

$P (x) = - \frac{3 t a n x}{l o g _{e} ( c o s x )}$
$Q (x) = \frac{- t a n x}{( l o g _{e} ( c o s x ) ) ^{2}}$

Setting up Integrating Factor

$I F = e^{\int P (x) d x} = e^{\int - \frac{3 t a n x}{l o g _{e} ( c o s x )} d x}$
Let $t = lo g_{e} (cos x) ⟹ d t = - tan x d x$

Calculating the Integrating Factor

$\int - \frac{3 t a n x}{l o g _{e} ( c o s x )} d x = \int \frac{3}{t} d t = 3 lo g_{e} t = lo g_{e} (t^{3})$
$I F = e^{l o g_{e} (t^{3})} = t^{3} = (lo g_{e} (cos x))^{3}$

The General Solution Formula

General Solution: $y \cdot I F = \int Q (x) \cdot I F d x + C$
Substitute values: $y (lo g_{e} (cos x))^{3} = \int \frac{- t a n x}{( l o g _{e} ( c o s x ) ) ^{2}} (lo g_{e} (cos x))^{3} d x + C$

Solving the RHS Integral

$y (lo g_{e} (cos x))^{3} = \int lo g_{e} (cos x) (- tan x) d x + C$
Let $t = lo g_{e} (cos x)$ , then $\int t d t = \frac{t ^{2}}{2} + C$
$y (lo g_{e} (cos x))^{3} = \frac{( l o g _{e} ( c o s x ) ) ^{2}}{2} + C$

Applying Initial Condition

Initial condition: $y (\frac{π}{4}) = - \frac{1}{l o g _{e} 2}$
At $x = \frac{π}{4}$ , $lo g_{e} (cos x) = lo g_{e} (\frac{1}{2}) = - \frac{1}{2} lo g_{e} 2$

Solving for Constant $C$

$(- \frac{1}{l o g _{e} 2}) (- \frac{1}{2} lo g_{e} 2)^{3} = \frac{( - \frac{1}{2} l o g _{e} 2 ) ^{2}}{2} + C$
$(- \frac{1}{l o g _{e} 2}) (- \frac{1}{8} (lo g_{e} 2)^{3}) = \frac{1}{8} (lo g_{e} 2)^{2} + C$
$\frac{1}{8} (lo g_{e} 2)^{2} = \frac{1}{8} (lo g_{e} 2)^{2} + C ⟹ C = 0$

The Particular Solution

$y (lo g_{e} (cos x))^{3} = \frac{( l o g _{e} ( c o s x ) ) ^{2}}{2}$
$y = \frac{1}{2 l o g _{e} ( c o s x )}$

Final Calculation for $y (\frac{π}{6})$

At $x = \frac{π}{6}$ , $cos x = \frac{3}{2}$
$y (\frac{π}{6}) = \frac{1}{2 l o g _{e} ( \frac{3}{2} )} = \frac{1}{l o g _{e} (( \frac{3}{2} ) ^{2} )}$

Simplifying the Final Answer

$y (\frac{π}{6}) = \frac{1}{l o g _{e} ( \frac{3}{4} )}$
Using $lo g_{e} (\frac{a}{b}) = lo g_{e} a - lo g_{e} b$ :
Final Answer: $\frac{1}{l o g _{e} 3 - l o g _{e} 4}$

Conclusion and Key Takeaways

Key Takeaway: Always look for a linear structure in complex differential equations.
Next Challenge: Try solving the same equation if the initial condition was $y (\frac{π}{3}) = 1$ .

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The Sigma Insight: Linear Differential Equations

Analyzing the Setup

The given differential equation is:

cos x (ln (cos x))^{2} d y + (sin x - 3 y sin x ln (cos x)) d x = 0

To solve this, we must transform it into the standard linear form:

\frac{d y}{d x} + P (x) y = Q (x)

Dividing the entire equation by

d x

and then by the coefficient of

d y

, which is

cos x (ln (cos x))^{2}

, we obtain:

\frac{d y}{d x} - \frac{3 tan x}{ln ( cos x )} y = \frac{- tan x}{( ln ( cos x ) ) ^{2}}

The Integrating Factor

We identify

P (x) = - \frac{3 t a n x}{l n ( c o s x )}

. The Integrating Factor (IF) is defined as:

I F = e^{\int P (x) d x} = e^{\int - \frac{3 t a n x}{l n ( c o s x )} d x}

Let

t = ln (cos x)

, then

d t = - tan x d x

. The integral simplifies significantly:

\int \frac{3}{t} d t = 3 ln ∣ t ∣

Exponentiating this result, we find:

I F = e^{3 l n (l n (c o s x))} = (ln (cos x))^{3}

The General Solution

The general solution is given by

y \cdot I F = \int Q (x) \cdot I F d x + C

. Substituting our values:

y (ln (cos x))^{3} = \int \frac{- tan x}{( ln ( cos x ) ) ^{2}} \cdot (ln (cos x))^{3} d x + C

Simplifying the integrand, we get:

y (ln (cos x))^{3} = \int - tan x ln (cos x) d x + C

Using the same substitution

t = ln (cos x)

, the integral becomes

\int t d t = \frac{t ^{2}}{2}

. Thus:

y (ln (cos x))^{3} = \frac{( ln ( cos x ) ) ^{2}}{2} + C

Applying Initial Conditions

We are given the condition

y (\frac{π}{4}) = - \frac{1}{l n 2}

. At

x = \frac{π}{4}

, we have:

ln (cos \frac{π}{4}) = ln (\frac{1}{2}) = - \frac{1}{2} ln 2

Substituting these into our general solution:

(- \frac{1}{ln 2}) (- \frac{1}{2} ln 2)^{3} = \frac{( - \frac{1}{2} ln 2 ) ^{2}}{2} + C

(- \frac{1}{ln 2}) (- \frac{1}{8} (ln 2)^{3}) = \frac{\frac{1}{4} ( ln 2 ) ^{2}}{2} + C

\frac{1}{8} (ln 2)^{2} = \frac{1}{8} (ln 2)^{2} + C \Rightarrow C = 0

Final Calculation

With

C = 0

, the particular solution is:

y = \frac{1}{2 ln ( cos x )}

To find

y (\frac{π}{6})

, we note

cos (\frac{π}{6}) = \frac{3}{2}

y = \frac{1}{2 ln ( \frac{3}{2} )} = \frac{1}{ln (( \frac{3}{2} ) ^{2} )} = \frac{1}{ln ( \frac{3}{4} )}

Applying logarithmic properties, the final answer is:

\frac{1}{ln 3 - ln 4}

Common Traps & FAQs

Wait, why can't we just separate the variables

x

and

y

and integrate directly?

That is a great question! Separation of variables only works if you can write the equation as

f (y) d y = g (x) d x

. Here, we have a term involving

y

multiplied by

sin x

and

ln (cos x)

, and another term involving

d y

. Because

y

is tied to

x

in a linear way, separation is impossible. We must use the linear differential equation method.

How did you know to substitute

t = ln (cos x)

? I never see these substitutions.

It comes from practice! Look at the coefficient of

y

, which is

P (x) = - \frac{3 t a n x}{l n ( c o s x )}

. The numerator is

- tan x

, which is exactly the derivative of

ln (cos x)

. Whenever you see a function and its derivative in the same expression, that is a huge signal to use substitution.

Why is the constant

C

equal to zero? Did I make a calculation error?

Not at all! It is perfectly normal for the constant of integration to be zero. When you plug in the initial condition

x = \frac{π}{4}

and

y = - \frac{1}{l n 2}

, the equation balances out perfectly such that

C

must be zero to satisfy the equality. Trust your algebra!

Select Answer:

Visualized Solution

Analyze the Differential Equation

Rearranging to Standard Form

Simplifying Coefficients

Identifying $P (x)$ and $Q (x)$

Setting up Integrating Factor

Calculating the Integrating Factor

The General Solution Formula

Solving the RHS Integral

Applying Initial Condition

Solving for Constant $C$

The Particular Solution

Final Calculation for $y (\frac{π}{6})$

Simplifying the Final Answer

Conclusion and Key Takeaways

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

The Integrating Factor

The General Solution

Applying Initial Conditions

Final Calculation

Similar Questions

If $\frac{d y}{d x} + \frac{3}{c o s ^{2} x} y = \frac{1}{c o s ^{2} x}, x \in (- \frac{π}{3}, \frac{π}{3})$ , and $y (\frac{π}{4}) = \frac{4}{3}$ , then $y (- \frac{π}{4})$ equals :

Let $y = y (x)$ be the solution of the differential equation $cos x (3 sin x + cos x + 3) d y = (1 + y sin x (3 sin x + cos x + 3)) d x$ , $0 \leq x \leq \frac{π}{2}, y (0) = 0$ . Then $y (\frac{π}{3})$ is equal to:

Let $y = y (x)$ be the solution of the differential equation $\frac{d y}{d x} = \frac{( t a n x ) + y}{s i n x ( s e c x - s i n x t an x )}, x \in (0, \frac{π}{2})$ satisfying the condition $y (\frac{π}{4}) = 2$ . Then, $y (\frac{π}{3})$ is

If $y = y (x), x \in (0, \frac{π}{2})$ be the solution curve of the differential equation $(sin^{2} 2 x) \frac{d y}{d x} + (8 sin^{2} 2 x + 2 sin 4 x) y = 2 e^{- 4 x} (2 sin 2 x + cos 2 x)$ , with $y (\frac{π}{4}) = e^{- π}$ , then $y (\frac{π}{6})$ is equal to :

Let $y = y (x)$ be the solution of the differential equation $2 cos x \frac{d y}{d x} = sin 2 x - 4 y sin x, x \in (0, \frac{π}{2}) .$ If $y (\frac{π}{3}) = 0$ then $y^{'} (\frac{π}{4}) + y (\frac{π}{4})$ is equal to

Let $y = y (x)$ be the solution of the differential equation $cos x \frac{d y}{d x} + 2 y sin x = sin 2 x$ , $x \in (0, \frac{π}{2})$ . If $y (π /3) = 0$ , then $y (π /4)$ is equal to :

If $y = y (x)$ is the solution of the differential equation $\frac{d y}{d x} + (tan x) y = sin x, 0 \leq x \leq \frac{π}{3}$ , with $y (0) = 0$ , then $y (\frac{π}{4})$ equal to :

Let $y = y (x)$ be the solution of the differential equation $(2 x lo g_{e} x) \frac{d y}{d x} + 2 y = \frac{3}{x} lo g_{e} x, x > 0$ and $y (e^{- 1}) = 0$ . Then, $y (e)$ is equal to

Let $y = y (x)$ be the solution of the differential equation $sin x \frac{d y}{d x} + y cos x = 4 x, x \in (0, π)$ . If $y (π /2) = 0$ , then $y (π /6)$ is equal to :

Let $y = y (x)$ be the solution curve of the differential equation $sin (2 x^{2}) lo g_{e} (tan x^{2}) d y + (4 x y - 42 x sin (x^{2} - \frac{π}{4})) d x = 0, 0 < x < \frac{π}{2}$ , which passes through the point $(\frac{π}{6}, 1)$ . Then $∣ y (\frac{π}{3}) ∣$ is equal to

Select Answer:

Visualized Solution

Analyze the Differential Equation

Rearranging to Standard Form

Simplifying Coefficients

Identifying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(x) and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q(x)

Setting up Integrating Factor

Calculating the Integrating Factor

The General Solution Formula

Solving the RHS Integral

Applying Initial Condition

Solving for Constant .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C

The Particular Solution

Final Calculation for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y(6π​)

Simplifying the Final Answer

Conclusion and Key Takeaways

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

The Integrating Factor

The General Solution

Applying Initial Conditions

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If dxdy​+cos2x3​y=cos2x1​,x∈(−3π​,3π​), and y(4π​)=34​, then y(−4π​) equals :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let y=y(x) be the solution of the differential equation cosx(3sinx+cosx+3)dy=(1+ysinx(3sinx+cosx+3))dx, 0≤x≤2π​,y(0)=0. Then y(3π​) is equal to:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let y=y(x) be the solution of the differential equation dxdy​=sinx(secx−sinx tanx)(tanx)+y​,x∈(0,2π​) satisfying the condition y(4π​)=2. Then, y(3π​) is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let y=y(x) be the solution of the differential equation 2 cos xdxdy​=sin 2x−4y sin x,x∈(0,2π​). If y(3π​)=0 then y′(4π​)+y(4π​) is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let y=y(x) be the solution of the differential equation cosxdxdy​+2ysinx=sin2x, x∈(0,2π​). If y(π/3)=0, then y(π/4) is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If y=y(x) is the solution of the differential equation dxdy​+(tanx)y=sinx,0≤x≤3π​, with y(0)=0, then y(4π​) equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let y=y(x) be the solution of the differential equation (2xloge​x)dxdy​+2y=x3​loge​x,x>0 and y(e−1)=0. Then, y(e) is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let y=y(x) be the solution of the differential equation sinxdxdy​+ycosx=4x,x∈(0,π). If y(π/2)=0, then y(π/6) is equal to :

Identifying $P (x)$ and $Q (x)$

Solving for Constant $C$

Final Calculation for $y (\frac{π}{6})$

If $\frac{d y}{d x} + \frac{3}{c o s ^{2} x} y = \frac{1}{c o s ^{2} x}, x \in (- \frac{π}{3}, \frac{π}{3})$ , and $y (\frac{π}{4}) = \frac{4}{3}$ , then $y (- \frac{π}{4})$ equals :

Let $y = y (x)$ be the solution of the differential equation $cos x (3 sin x + cos x + 3) d y = (1 + y sin x (3 sin x + cos x + 3)) d x$ , $0 \leq x \leq \frac{π}{2}, y (0) = 0$ . Then $y (\frac{π}{3})$ is equal to:

Let $y = y (x)$ be the solution of the differential equation $\frac{d y}{d x} = \frac{( t a n x ) + y}{s i n x ( s e c x - s i n x t an x )}, x \in (0, \frac{π}{2})$ satisfying the condition $y (\frac{π}{4}) = 2$ . Then, $y (\frac{π}{3})$ is

Let $y = y (x)$ be the solution of the differential equation $2 cos x \frac{d y}{d x} = sin 2 x - 4 y sin x, x \in (0, \frac{π}{2}) .$ If $y (\frac{π}{3}) = 0$ then $y^{'} (\frac{π}{4}) + y (\frac{π}{4})$ is equal to

Let $y = y (x)$ be the solution of the differential equation $cos x \frac{d y}{d x} + 2 y sin x = sin 2 x$ , $x \in (0, \frac{π}{2})$ . If $y (π /3) = 0$ , then $y (π /4)$ is equal to :

If $y = y (x)$ is the solution of the differential equation $\frac{d y}{d x} + (tan x) y = sin x, 0 \leq x \leq \frac{π}{3}$ , with $y (0) = 0$ , then $y (\frac{π}{4})$ equal to :

Let $y = y (x)$ be the solution of the differential equation $(2 x lo g_{e} x) \frac{d y}{d x} + 2 y = \frac{3}{x} lo g_{e} x, x > 0$ and $y (e^{- 1}) = 0$ . Then, $y (e)$ is equal to

Let $y = y (x)$ be the solution of the differential equation $sin x \frac{d y}{d x} + y cos x = 4 x, x \in (0, π)$ . If $y (π /2) = 0$ , then $y (π /6)$ is equal to :