Solve the differential equation $$rac{y dy}{dx} = x + xy^2$$ given that $y = 0$ when $x = 0$. The acceleration of a cyclist freewheeling down a slight hill is $$0.12 - 0.0006y^2 ext{ ms}^{-2}$$ where the velocity $y$ is in metres per second. The cyclist starts from rest at the top of the hill. Find (i) the speed of the cyclist after travelling 120 m down the hill (ii) the time taken by the cyclist to travel the 120 m if his average speed is 2.65 ms$^{-1}$.

Leaving Cert Applied Maths Differential Equations: Solve the differential equation

Solve the differential equation $$rac{y dy}{dx} = x + xy^2$$ given that $y = 0$ when $x = 0$ - Leaving Cert Applied Maths - Question 10 - 2010

Question 10

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Solve the differential equation $$rac{y dy}{dx} = x + xy^2$$ given that $y = 0$ when $x = 0$. The acceleration of a cyclist freewheeling down a slight hill ... show full transcript

Worked Solution & Example Answer:Solve the differential equation $$rac{y dy}{dx} = x + xy^2$$ given that $y = 0$ when $x = 0$ - Leaving Cert Applied Maths - Question 10 - 2010

Step 1

Solve the differential equation

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Answer

To solve the equation $y \frac{dy}{dx} = x + xy^2$ , we will separate the variables.

Rewrite the equation:
$y \frac{dy}{dx} = x(1 + y^2)$
Separate variables:
$\frac{y}{1 + y^2} dy = x dx$
Integrate both sides:
$\int \frac{y}{1 + y^2} dy = \int x dx$
- The left-hand side:
  $\frac{1}{2} \ln(1 + y^2)$
- The right-hand side:
  $\frac{1}{2} x^2 + C$
Setting the initial condition (y = 0) when (x = 0):
- We find the constant (C = 0).
Thus, the integrated equation is:
$\frac{1}{2} \ln(1 + y^2) = \frac{1}{2} x^2$
Solving for y:
$1 + y^2 = e^{x^2}$
$y^2 = e^{x^2} - 1$
$y = \sqrt{e^{x^2} - 1$ .

Step 2

Find (i) the speed of the cyclist after travelling 120 m down the hill

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Answer

Given acceleration equation:
$\frac{dy}{dx} = 0.12 - 0.0006y^2$
We can separate variables to find the speed:

Separate variables:
$\frac{y}{0.12 - 0.0006y^2} dy = dx$
Integrate both sides:
- The left side using partial fractions.
After integrating and substituting the limits, we get to
$y = 5.18 ext{ ms}^{-1}$

Step 3

Find (ii) the time taken by the cyclist to travel the 120 m if his average speed is 2.65 ms$^{-1}$

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Answer

Using the formula for average speed:
$\text{Average Speed} = \frac{\text{Distance}}{\text{Time}}$
Here, Distance = 120 m and Average Speed = 2.65 ms $^{-1}$ .

Rearranging gives:
$\text{Time} = \frac{\text{Distance}}{\text{Average Speed}}$
$t = \frac{120}{2.65}$
Calculating gives:
$t = 45.3 ext{ s}$

Solve the differential equation $$ rac{y dy}{dx} = x + xy^2$$ given that $y = 0$ when $x = 0$ - Leaving Cert Applied Maths - Question 10 - 2010

Worked Solution & Example Answer:Solve the differential equation $$ rac{y dy}{dx} = x + xy^2$$ given that $y = 0$ when $x = 0$ - Leaving Cert Applied Maths - Question 10 - 2010

Solve the differential equation

Find (i) the speed of the cyclist after travelling 120 m down the hill

Find (ii) the time taken by the cyclist to travel the 120 m if his average speed is 2.65 ms$^{-1}$

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Solve the differential equation $$rac{y dy}{dx} = x + xy^2$$ given that $y = 0$ when $x = 0$ - Leaving Cert Applied Maths - Question 10 - 2010

Worked Solution & Example Answer:Solve the differential equation $$rac{y dy}{dx} = x + xy^2$$ given that $y = 0$ when $x = 0$ - Leaving Cert Applied Maths - Question 10 - 2010