Question 28 of 96intermediate🔧 ApplyLong Answer4 marks

For each of the differential equations in Exercises 1 to 10, find the general solution:

sec⁡2xtan⁡y dx+sec⁡2ytan⁡x dy=0\sec^2 x \tan y \, dx + \sec^2 y \tan x \, dy = 0
Correct Answer

The given differential equation is:

sec⁡2xtan⁡y dx+sec⁡2ytan⁡x dy=0\sec^2 x \tan y \, dx + \sec^2 y \tan x \, dy = 0

To find the general solution, we first separate the variables xx and yy.

Rearranging the terms, we get:

sec⁡2xtan⁡y dx=−sec⁡2ytan⁡x dy\sec^2 x \tan y \, dx = - \sec^2 y \tan x \, dy

Dividing both sides by tan⁡xtan⁡y\tan x \tan y, we obtain:

sec⁡2xtan⁡x dx=−sec⁡2ytan⁡y dy\frac{\sec^2 x}{\tan x} \, dx = - \frac{\sec^2 y}{\tan y} \, dy

Integrating both sides with respect to xx and yy respectively:

∫sec⁡2xtan⁡x dx=−∫sec⁡2ytan⁡y dy\int \frac{\sec^2 x}{\tan x} \, dx = - \int \frac{\sec^2 y}{\tan y} \, dy

Using the substitution method, let tan⁡x=t\tan x = t, which implies sec⁡2x dx=dt\sec^2 x \, dx = dt. The integral becomes:

∫dtt=log⁡∣t∣=log⁡∣tan⁡x∣\int \frac{dt}{t} = \log|t| = \log|\tan x|

Similarly, for yy, the integral is log⁡∣tan⁡y∣\log|\tan y|.

Substituting these values into the integrated equation:

log⁡∣tan⁡x∣=−log⁡∣tan⁡y∣+log⁡C\log|\tan x| = - \log|\tan y| + \log C

Rearranging the logarithmic terms:

log⁡∣tan⁡x∣+log⁡∣tan⁡y∣=log⁡C\log|\tan x| + \log|\tan y| = \log C log⁡(∣tan⁡x∣⋅∣tan⁡y∣)=log⁡C\log(|\tan x| \cdot |\tan y|) = \log C

Taking the antilogarithm on both sides:

∣tan⁡x∣⋅∣tan⁡y∣=C|\tan x| \cdot |\tan y| = C

Thus, the general solution is:

tan⁡xtan⁡y=C\tan x \tan y = C
Exercise: EXERCISE 9.3 | Q: 4 | (Page 12)
For More Understanding

Explanation

This question asks for the general solution of a differential equation listed in the provided context (Question 4). The equation is of the variable separable form. The solution involves rearranging terms to group functions of xx with dxdx and functions of yy with dydy, followed by integration. The integral of sec⁡2(u)/tan⁡(u)\sec^2(u)/\tan(u) is a standard logarithmic form derived using substitution.

Solution Steps

  1. Step 1: Rewrite the given differential equation sec⁡2xtan⁡y dx+sec⁡2ytan⁡x dy=0\sec^2 x \tan y \, dx + \sec^2 y \tan x \, dy = 0 to separate variables:

    sec⁡2xtan⁡x dx=−sec⁡2ytan⁡y dy\frac{\sec^2 x}{\tan x} \, dx = - \frac{\sec^2 y}{\tan y} \, dy

  2. Step 2: Integrate both sides:

    ∫sec⁡2xtan⁡x dx=−∫sec⁡2ytan⁡y dy\int \frac{\sec^2 x}{\tan x} \, dx = - \int \frac{\sec^2 y}{\tan y} \, dy

  3. Step 3: Apply the substitution t=tan⁡xt = \tan x for the LHS and similarly for the RHS to get:

    log⁡∣tan⁡x∣=−log⁡∣tan⁡y∣+log⁡C\log|\tan x| = - \log|\tan y| + \log C

  4. Step 4: Combine the logarithmic terms:

    log⁡∣tan⁡x∣+log⁡∣tan⁡y∣=log⁡C\log|\tan x| + \log|\tan y| = \log C

    which simplifies to:

    log⁡(∣tan⁡xtan⁡y∣)=log⁡C\log(|\tan x \tan y|) = \log C

  5. Step 5: Apply the antilogarithm to find the final solution:

    tan⁡xtan⁡y=C\tan x \tan y = C
PreviousNext