Maths Q20 – JEE-MAIN 2025

Home Maths Differential EquationsQuestion #185

If a curve

y = y (x)

passes through the point

(1, \frac{π}{2})

and satisfies the differential equation

(7 x^{4} \cot y - e^{x} cosec y) \frac{d x}{d y} = x^{5}

x \geq 1

, then at

x = 2

, the value of

\cos y

is :

Clue 1 of 3: Differential Equation Transformation

The given differential equation is not in a standard linear form. Try to rearrange it to a more recognizable form, possibly by taking the reciprocal or making a substitution.

🧠 Full Solution Path

Step 1: Transforming the Differential Equation✦ Active

The given differential equation is $(7 x^{4} \cot y - e^{x} cosec y) \frac{d x}{d y} = x^{5}$ . We can rewrite it as $\frac{d y}{d x} = \frac{7 x^{4} \cot y - e^{x} cosec y}{x^{5}}$ . Dividing by $x^{5}$ and then by $cosec y$ (or multiplying by $\sin y$ ) gives:

\frac{d y}{d x} = \frac{7}{x} \cot y - \frac{e^{x}}{x^{5}} cosec y

\sin y \frac{d y}{d x} = \frac{7}{x} \frac{\cos y}{\sin y} \sin y - \frac{e^{x}}{x^{5}} \frac{1}{\sin y} \sin y

\sin y \frac{d y}{d x} - \frac{7}{x} \cos y = - \frac{e^{x}}{x^{5}}

Let $v = \cos y$ . Then $\frac{d v}{d x} = - \sin y \frac{d y}{d x}$ . Substituting this into the equation:

- \frac{d v}{d x} - \frac{7}{x} v = - \frac{e^{x}}{x^{5}}

\frac{d v}{d x} + \frac{7}{x} v = \frac{e^{x}}{x^{5}}

This is a linear first-order differential equation of the form $\frac{d v}{d x} + P (x) v = Q (x)$ , where $P (x) = \frac{7}{x}$ and $Q (x) = \frac{e^{x}}{x^{5}}$ .

Step 2: Solving the Linear Differential Equation○ Expand

The integrating factor (IF) is $e^{\int P (x) d x}$ :

I F = e^{\int \frac{7}{x} d x} = e^{7 \ln | x |} = e^{\ln (x^{7})} = x^{7} (since x \geq 1)

The general solution is $v \cdot I F = \int Q (x) \cdot I F d x + C$ :

v x^{7} = \int \frac{e^{x}}{x^{5}} \cdot x^{7} d x + C

v x^{7} = \int x^{2} e^{x} d x + C

Using integration by parts for $\int x^{2} e^{x} d x$ :

\int x^{2} e^{x} d x = x^{2} e^{x} - \int 2 x e^{x} d x = x^{2} e^{x} - (2 x e^{x} - \int 2 e^{x} d x) = x^{2} e^{x} - 2 x e^{x} + 2 e^{x} = e^{x} (x^{2} - 2 x + 2)

So, the solution is:

v x^{7} = e^{x} (x^{2} - 2 x + 2) + C

Substitute back $v = \cos y$ :

\cos y \cdot x^{7} = e^{x} (x^{2} - 2 x + 2) + C

💡 Teacher's Secret Hint

Remember to apply integration by parts carefully for the integral of $x^{2} e^{x}$ .

Step 3: Applying Initial Condition and Finding Value at x=2○ Expand

The curve passes through $(1, \frac{π}{2})$ . Substitute $x = 1$ and $y = \frac{π}{2}$ to find $C$ :

\cos (\frac{π}{2}) \cdot 1^{7} = e^{1} (1^{2} - 2 (1) + 2) + C

0 \cdot 1 = e (1 - 2 + 2) + C

0 = e + C ⟹ C = - e

The particular solution is:

\cos y \cdot x^{7} = e^{x} (x^{2} - 2 x + 2) - e

Now, find the value of $\cos y$ at $x = 2$ :

\cos y \cdot 2^{7} = e^{2} (2^{2} - 2 (2) + 2) - e

\cos y \cdot 128 = e^{2} (4 - 4 + 2) - e

\cos y \cdot 128 = 2 e^{2} - e

\cos y = \frac{2 e^{2} - e}{128}

💡 Teacher's Secret Hint

Ensure correct substitution of the initial point and careful calculation of the constant C.

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Maths Question 20 – JEE-MAIN 2025

Clue 1 of 3: Differential Equation Transformation

🧠 Full Solution Path