Question

Show that the tangents to all integral curves of the differential equation \(y^{\prime}+y \tan x=x \tan x+1\) at the points of intersection with the y-axis are parallel. Determine the angle at which the integral curves cut the \(\mathrm{y}\)-axis.

Short answer

Question: Show that the tangents to all integral curves of the differential equation \(y^\prime + y\tan(x) = \tan(x)\) at the points of intersection with the y-axis are parallel and determine the angle at which the integral curves cut the y-axis. Answer: The tangents to all integral curves at the points of intersection with the y-axis are parallel, with a constant slope of 1. The angle at which the integral curves cut the y-axis is \(\frac{\pi}{4}\) radians (or \(45^{\circ}\)).

Step-by-step solution

Calculate the slope of the tangent at the points of intersection with the y-axis

First, find the points where the integral curves intersect the y-axis. The y-axis is given by \(x = 0\), so substitute this value into the given differential equation: \(y^\prime + y \tan(0) = 0 \tan(0) + 1\) Since \(\tan(0) = 0\), this simplifies to: \(y^\prime = 1\) The slope of the tangent at the points of intersection with the y-axis is equal to the derivative \(y^\prime\) evaluated at \(x=0\). Since we have just shown that \(y^\prime = 1\) for all integral curves at this point, the slope is constant and equal to 1.

Show that the slopes of the tangents are equal for all integral curves

In step 1, we calculated the slope of the tangent at the points of intersection with the y-axis for all integral curves to be 1. Since all tangents have a constant slope of 1, we have shown that the tangents to all integral curves of the differential equation are parallel at the points of intersection with the y-axis.

Determine the angle at which the integral curves cut the y-axis

Since the integral curves are parallel to the tangents, and we know the slopes of the tangents are equal to 1, we can use the tangent function to find the angle at which they cut the y-axis. The tangent function relates the slope of the tangent to the angle, so we have: \(\tan(\theta) = 1\) Now, find the angle \(\theta\): \(\theta = \arctan(1)\) Since the angle whose tangent is 1 is \(\frac{\pi}{4}\) radians (or \(45^{\circ}\)), we have: \(\theta = \frac{\pi}{4}\) radians (or \(45^{\circ}\)) Hence, the angle at which the integral curves cut the y-axis is \(\frac{\pi}{4}\) radians (or \(45^{\circ}\)).