If $$x^2 \frac{dy}{dx} - xy = 7y$$ and $y = 1$ when $x = 1$, find the value of $y$ when $x = 2$. (b) A particle travelling in a straight line has a deceleration of $$\frac{v^2}{400} + 16 \text{ ms}^{-2}$$ where $v$ is its speed at any time $t$. If its initial speed is 40 ms$^{-1}$, find (i) the distance travelled before it comes to rest (ii) the average speed of the particle during the motion.

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If $$x^2 \frac{dy}{dx} - xy = 7y$$ and $y = 1$ when $x = 1$, find the value of $y$ when $x = 2$ - Leaving Cert Applied Maths - Question 10 - 2011

Question 10

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If $$x^2 \frac{dy}{dx} - xy = 7y$$ and $y = 1$ when $x = 1$, find the value of $y$ when $x = 2$. (b) A particle travelling in a straight line has a deceleratio... show full transcript

Worked Solution & Example Answer:If $$x^2 \frac{dy}{dx} - xy = 7y$$ and $y = 1$ when $x = 1$, find the value of $y$ when $x = 2$ - Leaving Cert Applied Maths - Question 10 - 2011

Step 1

Find the value of $y$ when $x = 2$

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Answer

We start with the equation:

$x^2 \frac{dy}{dx} - xy = 7y$

Rearranging gives:

$\frac{dy}{dx} = \frac{7y + xy}{x^2}$

Separate Variables:

Rearranging allows us to separate the variables:

$\frac{dy}{y} = \frac{7 + \frac{x}{x^2}}{x^2} dx$

Integrate:

Integrating both sides, we have:

$\int \frac{1}{y} dy = \int \left( \frac{7}{x^2} + \frac{1}{x} \right) dx$

The integrals yield:

$\ln |y| = -\frac{7}{x} + \ln |x| + C$

Find the Constant C:

Using the condition $y = 1$ when $x = 1$ :

$\ln(1) = -7 + \ln(1) + C \Rightarrow C = 7$

Substitute C Back:

Thus, we have:

$\ln |y| = -\frac{7}{x} + \ln |x| + 7$

Calculate for $x = 2$ :

At $x = 2$ :

\ln |y| = -\frac{7}{2} + \ln(2) + 7

Solving this gives:

$y = e^{\ln(2) + 7 - 3.5} = e^{4.1931} \approx 66.23$

Step 2

the distance travelled before it comes to rest

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Answer

To find the distance travelled before the particle comes to rest, we use the equation of motion with deceleration:

Rewrite the Deceleration Equation:

$\frac{dv}{dt} = -\left( \frac{v^2}{400} + 16 \right)$

Separate Variables:

This can be written as:

$\frac{dv}{\frac{v^2}{400} + 80} = -dt$

Integrate:

Integrating gives:

$\int_0^{40} \frac{1}{\frac{v^2}{400} + 80} dv = -\int_0^T dt$

Calculate Limits:

Using the substitution, we find the distance:

$x = 200ln\left(\frac{40 + 80}{80}\right) = 200\ln\left(\frac{3}{2}\right)\approx 44.63 ext{ m}$

Step 3

the average speed of the particle during the motion

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Answer

To find the average speed:

Use Distance and Time:

Average speed is given by:

$\text{average speed} = \frac{\text{total distance}}{\text{total time}}$

Apply Values:

From our previous calculations:

Total distance = 44.63 m

Total time = 2.32 s

Final Calculation:

Thus, the average speed is:

$\text{average speed} = \frac{44.63}{2.32} \approx 19.24 \text{ m s}^{-1}$

If $$x^2 \frac{dy}{dx} - xy = 7y$$ and $y = 1$ when $x = 1$, find the value of $y$ when $x = 2$ - Leaving Cert Applied Maths - Question 10 - 2011

Worked Solution & Example Answer:If $$x^2 \frac{dy}{dx} - xy = 7y$$ and $y = 1$ when $x = 1$, find the value of $y$ when $x = 2$ - Leaving Cert Applied Maths - Question 10 - 2011

Find the value of $y$ when $x = 2$

the distance travelled before it comes to rest

the average speed of the particle during the motion

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