Question

If y=f(x) is the solution of the differential equation \( (1+\cos^2 x)f'(x) - 4\sin(2x) - f(x)\sin(2x) = 0 \) when f(0)=0, then \( f(\pi/3) = \)

• A. 3
• B. \( \frac{12}{5} \)
• C. \( \frac{3}{5} \)
• D. 4

Hint:

When solving a first-order differential equation, first try to determine if it is separable. A separable equation can be written in the form \(g(y)dy = h(x)dx\), allowing you to integrate both sides independently.

Answer 1

The Correct Option is B

Let's rearrange the differential equation. Let \(y=f(x)\).
\( (1+\cos^2 x)\frac{dy}{dx} = 4\sin(2x) + y\sin(2x) = \sin(2x)(4+y) \).
This is a separable differential equation.
\( \frac{dy}{4+y} = \frac{\sin(2x)}{1+\cos^2 x} dx \).
Integrate both sides.
\( \int \frac{dy}{4+y} = \ln(4+y) \). (Assuming \(4+y>0\)).
For the RHS: \( \int \frac{2\sin x \cos x}{1+\cos^2 x} dx \).
Let \( u = 1+\cos^2 x \). Then \( du = 2\cos x (-\sin x) dx = -\sin(2x) dx \).
So the RHS integral is \( \int \frac{-du}{u} = -\ln|u| = -\ln(1+\cos^2 x) \).
Combining the results: \( \ln(4+y) = -\ln(1+\cos^2 x) + C \).
\( \ln(4+y) + \ln(1+\cos^2 x) = C \implies \ln((4+y)(1+\cos^2 x)) = C \).
\( (4+y)(1+\cos^2 x) = e^C = K \) (another constant).
Use the initial condition \(f(0)=0\) (i.e., \(y=0\) when \(x=0\)) to find K.
\( (4+0)(1+\cos^2 0) = K \implies (4)(1+1) = K \implies K=8 \).
The solution is \( (4+y)(1+\cos^2 x) = 8 \).
We need to find \(y = f(\pi/3)\).
\( (4+y)(1+\cos^2(\pi/3)) = 8 \).
\( \cos(\pi/3) = 1/2 \), so \( \cos^2(\pi/3) = 1/4 \).
\( (4+y)(1+1/4) = 8 \implies (4+y)(5/4) = 8 \).
\( 4+y = 8 \cdot (4/5) = 32/5 \).
\( y = \frac{32}{5} - 4 = \frac{32-20}{5} = \frac{12}{5} \).