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Find the particular solution to the differential equation \((x - 2) \frac{dy}{dx} = xy\) that passes through the point (0, 1) - HSC - SSCE Mathematics Extension 1 - Question 14 - 2022 - Paper 1
Question 14
Find the particular solution to the differential equation \((x - 2) \frac{dy}{dx} = xy\) that passes through the point (0, 1). The vectors \(\vec{a}\) and \(\vec{b}... show full transcript
Worked Solution & Example Answer:Find the particular solution to the differential equation \((x - 2) \frac{dy}{dx} = xy\) that passes through the point (0, 1) - HSC - SSCE Mathematics Extension 1 - Question 14 - 2022 - Paper 1
Step 1
Find the particular solution to the differential equation \((x - 2) \frac{dy}{dx} = xy\) that passes through the point (0, 1).
Answer
To solve the differential equation, we first rearrange it:
[ \frac{dy}{dx} = \frac{xy}{x - 2} ]
This is a separable differential equation, which we can rewrite as:
[ \frac{dy}{y} = \frac{x}{x - 2} dx ]
Integrating both sides gives:
[ \ln|y| = \int \frac{x}{x - 2} dx ]
Using partial fraction decomposition, we can express:
[ \frac{x}{x - 2} = 1 + \frac{2}{x - 2} ]
Thus, the integral becomes:
[ \ln|y| = x + 2 \ln|x - 2| + C ]
Exponentiating both sides, we find:
[ y = e^{C} \cdot (x - 2)^{2} \cdot e^{x} ]
Using the initial condition (y(0) = 1):
[ 1 = e^{C} \cdot (-2)^{2} ]
Thus, (C = \ln(\frac{1}{4})). Finally, substituting back gives:
[ y = \frac{(x - 2)^{2} e^{x}}{4} ]
Step 2
The vectors \(\vec{a}\) and \(\vec{b}\) are not parallel. The vector \(\vec{p}\) is the projection of \(\vec{b}\) onto the vector \(\vec{a}\).
Answer
To find the projection of (\vec{b}) onto (\vec{a}):
[ \vec{p} = \text{proj}{\vec{a}} \vec{b} = \frac{\vec{a} \cdot \vec{b}}{\vec{a} \cdot \vec{a}} \vec{a} ]
To show (\lvert \vec{a} - \lambda \vec{b} \rvert) is minimized when (\lambda = \lambda_0):
We differentiate (\lvert \vec{a} - \lambda \vec{b} \rvert^{2}) with respect to (\lambda) and set it to zero, getting:
[ \frac{d}{d\lambda}(\lvert \vec{a} - \lambda \vec{b} \rvert^{2}) = 0 ]
This leads to (\lambda = \lambda{0}) being the condition for minimizing the distance.
Step 3
Show that, for the player to have a chance of hitting the target, \(d\) must be less than 37% of the maximum possible range of the projectile (to 2 significant figures).
Answer
The maximum range (R) of a projectile is given by the formula:
[ R = \frac{u^2 \sin(2\theta)}{g} ]
For our scenario, the target moves away at (\frac{u}{2}). For the player to hit the target:
[ d < 0.37 R ]
Substituting yields:
[ d < 0.37 \cdot \frac{(2u)^2 \sin(2\theta)}{g} ]
Simplifying provides the necessary relation to determine the distance (d) in terms of projectile launch speed and angle.
Step 4
An airline company that has empty seats on a flight is not maximising its profit.
Answer
The company is dealing with overbooking and wants to manage passenger load effectively. Given the probability (P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}) for flight bookings, the constraint should be set to ensure that more than 10% of instances leads to overbooking. With a total of 350 seats and a probability of 0.05 for no-shows, we can set up the inequality to solve for maximum seats they can sell while keeping overbooked percentages under control.