Question

If $y=y(x)$, $y \in [0, \pi/2)$ is the solution of the differential equation $\sec y \frac{dy}{dx} - \sin(x+y) - \sin(x-y) = 0$, with $y(0)=0$, then $5y'(\pi/2)$ is equal to _________.

Hint:

When asked to find the value of a derivative at a point for an implicitly defined function, you don't always need to find the explicit function $y(x)$. Instead, use the differential equation itself, find the value of y at the required point, and then substitute everything into the DE to find the value of $y'$.

Answer 1

Correct Answer: 2

Step 1: Simplify the differential equation Using the identity \[ \sin(x+y)+\sin(x-y)=2\sin x\cos y \] the given equation becomes \[ \sec y\,\frac{dy}{dx}-2\sin x\cos y=0 \] \[ \frac{1}{\cos y}\frac{dy}{dx}=2\sin x\cos y \] \[ \frac{dy}{dx}=2\sin x\cos^2 y \] Step 2: Separate the variables \[ \frac{1}{\cos^2 y}\,dy=2\sin x\,dx \] \[ \sec^2 y\,dy=2\sin x\,dx \] Step 3: Integrate both sides \[ \int \sec^2 y\,dy=\int 2\sin x\,dx \] \[ \tan y=-2\cos x+C \] Step 4: Use the initial condition Given \(y(0)=0\), \[ \tan 0=-2\cos 0+C \] \[ 0=-2+C \Rightarrow C=2 \] Hence, \[ \tan y=2-2\cos x \] Step 5: Find \(y(\pi/2)\) \[ \tan y\Big|_{x=\pi/2}=2-2\cos\frac{\pi}{2}=2 \] \[ \tan y(\pi/2)=2 \] Using \[ 1+\tan^2 y=\sec^2 y \] \[ \sec^2 y=1+4=5 \Rightarrow \cos^2 y=\frac{1}{5} \] Step 6: Evaluate \(y'(\pi/2)\) From \[ \frac{dy}{dx}=2\sin x\cos^2 y \] \[ y'(\pi/2)=2\sin\frac{\pi}{2}\cdot\frac{1}{5}=\frac{2}{5} \] Step 7: Final value \[ 5y'(\pi/2)=5\times\frac{2}{5}=2 \] Answer: \(\boxed{2}\)