Question

Let \( y = y(x) \) be the solution of the differential equation \( \frac{dy}{dx} + 3(\tan^2 x) y + 3y = \sec^2 x \), with \( y(0) = \frac{1}{3} + e^3 \). Then \( y\left(\frac{\pi}{4}\right) \) is equal to

• A. \( \frac{2}{3} \)
• B. \( \frac{4}{3} \)
• C. \( \frac{4}{3} + e^3 \)
• D. \( \frac{2}{3} + e^3 \)

Hint:

Recognize the linear differential equation form \( \frac{dy}{dx} + P(x) y = Q(x) \). Find the integrating factor \( IF = e^{\int P(x) dx} \). The solution is \( y \cdot IF = \int Q(x) \cdot IF \, dx + c \). Use the initial condition to find the value of the constant \( c \), and then substitute the required value of \( x \) to find \( y \). Remember trigonometric identities to simplify the equation and integration.

Answer 1

The Correct Option is B

The given differential equation is: \[ \frac{dy}{dx} + 3(\tan^2 x) y + 3y = \sec^2 x \] \[ \frac{dy}{dx} + 3(\tan^2 x + 1) y = \sec^2 x \] Using the identity \( \tan^2 x + 1 = \sec^2 x \): \[ \frac{dy}{dx} + 3\sec^2 x \cdot y = \sec^2 x \] This is a linear differential equation of the form \( \frac{dy}{dx} + P(x) y = Q(x) \), where \( P(x) = 3\sec^2 x \) and \( Q(x) = \sec^2 x \). The integrating factor (IF) is given by \( e^{\int P(x) dx} \): \[ IF = e^{\int 3\sec^2 x dx} = e^{3\tan x} \] The solution of the linear differential equation is: \[ y \cdot IF = \int Q(x) \cdot IF \, dx + c \] \[ y \cdot e^{3\tan x} = \int \sec^2 x \cdot e^{3\tan x} \, dx + c \] Let \( u = 3\tan x \), then \( du = 3\sec^2 x \, dx \), so \( \sec^2 x \, dx = \frac{1}{3} du \). \[ y \cdot e^{3\tan x} = \int e^u \cdot \frac{1}{3} \, du + c \] \[ y \cdot e^{3\tan x} = \frac{1}{3} e^u + c \] Substitute back \( u = 3\tan x \): \[ y \cdot e^{3\tan x} = \frac{1}{3} e^{3\tan x} + c \] Given the initial condition \( y(0) = \frac{1}{3} + e^3 \). When \( x = 0 \), \( \tan(0) = 0 \). \[ \left(\frac{1}{3} + e^3\right) e^{3(0)} = \frac{1}{3} e^{3(0)} + c \] \[ \frac{1}{3} + e^3 = \frac{1}{3} (1) + c \] \[ \frac{1}{3} + e^3 = \frac{1}{3} + c \] \[ c = e^3 \] The particular solution is: \[ y \cdot e^{3\tan x} = \frac{1}{3} e^{3\tan x} + e^3 \] We need to find \( y\left(\frac{\pi}{4}\right) \). When \( x = \frac{\pi}{4} \), \( \tan\left(\frac{\pi}{4}\right) = 1 \). \[ y \cdot e^{3(1)} = \frac{1}{3} e^{3(1)} + e^3 \] \[ y \cdot e^3 = \frac{1}{3} e^3 + e^3 \] Divide by \( e^3 \): \[ y = \frac{1}{3} + 1 = \frac{4}{3} \] So, \( y\left(\frac{\pi}{4}\right) = \frac{4}{3} \).

Answer 2

To solve the given differential equation, we need to apply the integrating factor method. The differential equation provided is:

\(\frac{dy}{dx} + 3(\tan^2 x)y + 3y = \sec^2 x\)

Rewriting this equation, we get:

\(\frac{dy}{dx} + 3(1 + \tan^2 x)y = \sec^2 x\)

Here, \(3(1 + \tan^2 x) = 3\sec^2 x\). So, the equation becomes:

\(\frac{dy}{dx} + 3\sec^2 x \cdot y = \sec^2 x\)

Identifying this as a linear first-order differential equation in the standard form \(\frac{dy}{dx} + P(x)y = Q(x)\), where \(P(x) = 3\sec^2 x\) and \(Q(x) = \sec^2 x\).

The integrating factor (IF) is given by:

\(\text{IF} = e^{\int P(x) \, dx} = e^{\int 3\sec^2 x \, dx}\)

The integral of \(\sec^2 x\) is \(\tan x\), so:

\(\text{IF} = e^{3\tan x}\)

Multiplying the entire differential equation by this integrating factor, we have:

\(e^{3\tan x} \frac{dy}{dx} + 3\sec^2 x e^{3\tan x} y = \sec^2 x e^{3\tan x}\)

The left-hand side is a derivative of \(y \cdot e^{3\tan x}\). Therefore:

\(\frac{d}{dx} \left( y \cdot e^{3\tan x} \right) = \sec^2 x e^{3\tan x}\)

Integrating both sides with respect to \(x\) gives:

\(y \cdot e^{3\tan x} = \int \sec^2 x e^{3\tan x} \, dx + C\)

We already know the formula for the integration by parts. From integration, we have:

\(\int e^{3\tan x} \sec^2 x \, dx = \frac{e^{3\tan x}}{3}\)

Thus, the solution becomes:

\(y \cdot e^{3\tan x} = \frac{e^{3\tan x}}{3} + C\)

Solving for \(y\) gives:

\(y = \frac{1}{3} + C e^{-3\tan x}\)

Using the initial condition \(y(0) = \frac{1}{3} + e^3\), we substitute \(x = 0\):

\(\frac{1}{3} + e^3 = \frac{1}{3} + C e^{0}\)

This gives us \(C = e^3\).

Thus, the particular solution is:

\(y = \frac{1}{3} + e^3 e^{-3\tan x}\)

Substitute \(x = \frac{\pi}{4}\):

\(y\left(\frac{\pi}{4}\right) = \frac{1}{3} + e^3 e^{-3}\bigg(\tan\left(\frac{\pi}{4}\right)\bigg)\)

Since \(\tan\left(\frac{\pi}{4}\right) = 1\), we have:

\(y\left(\frac{\pi}{4}\right) = \frac{1}{3} + e^3 e^{-3}\)

This results in:

\(y\left(\frac{\pi}{4}\right) = \frac{1}{3} + 1\)

Thus, \(y\left(\frac{\pi}{4}\right) = \frac{4}{3}\).

Therefore, the correct answer is: \(\frac{4}{3}\)