Question 67 of 96intermediate🔧 ApplyLong Answer4 marks

For each of the differential equations given in Exercises 1 to 12, find the general solution:

dydx+(sec⁡x)y=tan⁡x(0≤x<π2)\frac{dy}{dx} + (\sec x) y = \tan x \quad \left(0 \leq x < \frac{\pi}{2}\right)
Correct Answer

The given differential equation is:

dydx+(sec⁡x)y=tan⁡x\frac{dy}{dx} + (\sec x) y = \tan x

This is a linear differential equation of the form dydx+Py=Q\frac{dy}{dx} + Py = Q, where P=sec⁡xP = \sec x and Q=tan⁡xQ = \tan x.

First, we find the Integrating Factor (I.F.):

I.F.=e∫P dx=e∫sec⁡x dx=elog⁡(sec⁡x+tan⁡x)=sec⁡x+tan⁡xI.F. = e^{\int P \, dx} = e^{\int \sec x \, dx} = e^{\log(\sec x + \tan x)} = \sec x + \tan x

(Note: for 0≤x<π20 \leq x < \frac{\pi}{2}, sec⁡x+tan⁡x>0\sec x + \tan x > 0).

The solution of the linear differential equation is given by:

y(I.F.)=∫Q(I.F.) dx+Cy(I.F.) = \int Q(I.F.) \, dx + C

Substituting the values of I.F. and Q:

y(sec⁡x+tan⁡x)=∫tan⁡x (sec⁡x+tan⁡x) dx+Cy(\sec x + \tan x) = \int \tan x \, (\sec x + \tan x) \, dx + C

Now, we evaluate the integral on the right-hand side:

∫(tan⁡xsec⁡x+tan⁡2x) dx=∫(tan⁡xsec⁡x+sec⁡2x−1) dx\int (\tan x \sec x + \tan^2 x) \, dx = \int (\tan x \sec x + \sec^2 x - 1) \, dx

Integrating term by term:

=sec⁡x+tan⁡x−x= \sec x + \tan x - x

Substituting this back into the solution equation:

y(sec⁡x+tan⁡x)=sec⁡x+tan⁡x−x+Cy(\sec x + \tan x) = \sec x + \tan x - x + C

This is the required general solution.

Exercise: EXERCISE 9.5 | Q: 4 | (Page 29)
For More Understanding

Explanation

The question asks for the general solution of a linear differential equation listed in Exercise 9.5 of the NCERT textbook. The method involves calculating an integrating factor and integrating the product of the integrating factor and the non-homogeneous term.

The provided context confirms the question is from Exercise 9.5, where preceding questions (1-3) are also linear differential equations, validating the solution approach.

Solution Steps

  1. Step 1: Identify the equation as linear: dydx+Py=Q\frac{dy}{dx} + Py = Q with P=sec⁡xP = \sec x and Q=tan⁡xQ = \tan x.

  2. Step 2: Calculate the Integrating Factor:

    I.F.=e∫sec⁡x dx=sec⁡x+tan⁡xI.F. = e^{\int \sec x \, dx} = \sec x + \tan x

  3. Step 3: Multiply the equation by the I.F.:

    y(sec⁡x+tan⁡x)=∫tan⁡x (sec⁡x+tan⁡x) dx+Cy(\sec x + \tan x) = \int \tan x \, (\sec x + \tan x) \, dx + C

  4. Step 4: Simplify the integrand:

    ∫(tan⁡xsec⁡x+sec⁡2x−1) dx\int (\tan x \sec x + \sec^2 x - 1) \, dx

  5. Step 5: Integrate to get:

    sec⁡x+tan⁡x−x\sec x + \tan x - x

  6. Step 6: Write the final general solution:

    y(sec⁡x+tan⁡x)=sec⁡x+tan⁡x−x+Cy(\sec x + \tan x) = \sec x + \tan x - x + C
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