If $$rac{x^2 rac{dy}{dx}}{x^2} - 7 = 0$$ and $y = 1$ when $x = 7$, find the value of $y$ when $x = 14$. (b) A particle starts from rest at $O$ at time $t = 0$. It travels along a straight line with acceleration $(24 - 16)$ ms$^{-2}$, where $t$ is the time measured from the instant when the particle is at $O$. Find (i) its velocity and its distance from $O$ at time $t = 3$ (ii) the value of $t$ when the speed of the particle is $80$ ms$^{-1}$. (c) Water flows from a tank at a rate proportional to the volume of water remaining in the tank. The tank is initially full and after one hour it is half full. After how many more minutes will it be one-fifth full?

Leaving Cert Applied Maths Differential Equations: If $$rac{x^2 rac{dy}{dx}}{x^2

If $$rac{x^2 rac{dy}{dx}}{x^2} - 7 = 0$$ and $y = 1$ when $x = 7$, find the value of $y$ when $x = 14$ - Leaving Cert Applied Maths - Question 10 - 2013

Question 10

$If--$$-rac{x^2--rac{dy}{dx}}{x^2}---7-=-0$$-and-$y-=-1$-when-$x-=-7$,-find-the-value-of-$y$-when-$x-=-14$-Leaving Cert Applied Maths-Question 10-2013.png$

If $$rac{x^2 rac{dy}{dx}}{x^2} - 7 = 0$$ and $y = 1$ when $x = 7$, find the value of $y$ when $x = 14$. (b) A particle starts from rest at $O$ at time $t = 0$. I... show full transcript

Worked Solution & Example Answer:If $$rac{x^2 rac{dy}{dx}}{x^2} - 7 = 0$$ and $y = 1$ when $x = 7$, find the value of $y$ when $x = 14$ - Leaving Cert Applied Maths - Question 10 - 2013

Step 1

If $$rac{x^2 rac{dy}{dx}}{x^2} - 7 = 0$$ and $y = 1$ when $x = 7$, find the value of $y$ when $x = 14$.

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Answer

To solve the differential equation, we first rearrange it: $rac{dy}{dx} = 7$ Integrating both sides, we find: $y = 7x + C$ Using the condition $y = 1$ when $x = 7$ :

C = 1 - 49 = -48$$ Thus, the equation becomes: $$y = 7x - 48$$ Now, substituting $x = 14$: $$y = 7(14) - 48 = 98 - 48 = 50$$ Therefore, when $x = 14$, $y = 50$.

Step 2

(b) A particle starts from rest at $O$ at time $t = 0$... Find (i) its velocity and its distance from $O$ at time $t = 3$.

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Answer

The acceleration is given as $a = 24 - 16$ . Integrating this to find the velocity:

v = 8t + C$$ Since the particle starts from rest, $v(0) = 0 \ C = 0$, hence: v = 8t. Now calculating velocity at $t = 3$: $$v(3) = 8(3) = 24 \, \text{ms}^{-1}$$ For the distance: $$s = \int_0^t v \, dt = \int_0^3 (8t) \, dt = [4t^2]_0^3 = 4(3^2) - 0 = 36 (\text{m})$$

Step 3

(b) (ii) the value of $t$ when the speed of the particle is $80$ ms$^{-1}$.

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Answer

Setting the equation for velocity:

t = 10 \, \text{s}$$

Step 4

(c) Water flows from a tank at a rate proportional to the volume of water remaining in the tank. After how many more minutes will it be one-fifth full?

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Answer

Let the volume of water in the tank be $V$ . The rate of change of volume can be described by: $\frac{dV}{dt} = -kV$ Integrating gives: $ln |V| = -kt + C$ At $t=0$ , $V_0$ : $ln |V_0| = C$ , At $t=1$ , $V = \frac{1}{2}V_0$ :

hence, \ln(2) = k \ k = ln(2)$$ To find the time it will take to go from $ rac{1}{2}V_0$ to $ rac{1}{5}V_0$: $$ln |\frac{1}{5}| - ln |\frac{1}{2}| = -kt \ t = \frac{ln(2/5)}{\ln(2)}. \ Substituting gives approximately: t \approx 1.22 h = 79.3 \text{minutes}$$

If $$ rac{x^2 rac{dy}{dx}}{x^2} - 7 = 0$$ and $y = 1$ when $x = 7$, find the value of $y$ when $x = 14$ - Leaving Cert Applied Maths - Question 10 - 2013

Worked Solution & Example Answer:If $$ rac{x^2 rac{dy}{dx}}{x^2} - 7 = 0$$ and $y = 1$ when $x = 7$, find the value of $y$ when $x = 14$ - Leaving Cert Applied Maths - Question 10 - 2013

If $$ rac{x^2 rac{dy}{dx}}{x^2} - 7 = 0$$ and $y = 1$ when $x = 7$, find the value of $y$ when $x = 14$.

(b) A particle starts from rest at $O$ at time $t = 0$... Find (i) its velocity and its distance from $O$ at time $t = 3$.

(b) (ii) the value of $t$ when the speed of the particle is $80$ ms$^{-1}$.

(c) Water flows from a tank at a rate proportional to the volume of water remaining in the tank. After how many more minutes will it be one-fifth full?

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If $$rac{x^2 rac{dy}{dx}}{x^2} - 7 = 0$$ and $y = 1$ when $x = 7$, find the value of $y$ when $x = 14$ - Leaving Cert Applied Maths - Question 10 - 2013

Worked Solution & Example Answer:If $$rac{x^2 rac{dy}{dx}}{x^2} - 7 = 0$$ and $y = 1$ when $x = 7$, find the value of $y$ when $x = 14$ - Leaving Cert Applied Maths - Question 10 - 2013

If $$rac{x^2 rac{dy}{dx}}{x^2} - 7 = 0$$ and $y = 1$ when $x = 7$, find the value of $y$ when $x = 14$.