10. (a) Solve the differential equation $ \frac{dy}{dx} = y^3 \sin x $ given that $ y = 1 $ when $ x = \frac{\pi}{2} $. (b) The acceleration of a racing car at a speed of $ v $ m/s is $ \left( 1 - \frac{v^2}{3200} \right) $ m/s² The car starts from rest. Calculate correct to two decimal places (i) the speed of the car when it has travelled 1500 m from rest (ii) the maximum speed of the car.

Leaving Cert Applied Maths Differential Equations: 10. (a) Solve the differential

10. (a) Solve the differential equation $ \frac{dy}{dx} = y^3 \sin x $ given that $ y = 1 $ when $ x = \frac{\pi}{2} $ - Leaving Cert Applied Maths - Question 10 - 2007

Question 10

$10.-(a)-Solve-the-differential-equation-$-\frac{dy}{dx}-=-y^3-\sin-x-$--given-that-$-y-=-1-$-when-$-x-=-\frac{\pi}{2}-$-Leaving Cert Applied Maths-Question 10-2007.png$

10. (a) Solve the differential equation $ \frac{dy}{dx} = y^3 \sin x $ given that $ y = 1 $ when $ x = \frac{\pi}{2} $. (b) The acceleration of a racing ... show full transcript

Worked Solution & Example Answer:10. (a) Solve the differential equation $ \frac{dy}{dx} = y^3 \sin x $ given that $ y = 1 $ when $ x = \frac{\pi}{2} $ - Leaving Cert Applied Maths - Question 10 - 2007

Step 1

Solve the differential equation $ \frac{dy}{dx} = y^3 \sin x $

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Answer

To solve the differential equation, we first separate the variables:

$\frac{1}{y^3} dy = \sin x \, dx$

Integrating both sides gives:

$-\frac{1}{2y^2} = -\cos x + C$

Rearranging leads to:

$\frac{1}{y^2} = 2\cos x + C$

John substituting for the constant ( C ) using the condition ( y = 1 ) when ( x = \frac{\pi}{2} ):

( \frac{1}{1^2} = 2\cos(\frac{\pi}{2}) + C )\n( \Rightarrow 1 = 0 + C \Rightarrow C = 1 )

So,

$\frac{1}{y^2} = 2\cos x + 1$

Finally, we find:

$y = \frac{1}{1 + 2 \cos x}$

Step 2

Calculate correct to two decimal places the speed of the car when it has travelled 1500 m from rest

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Answer

To find the speed of the car after it has travelled 1500 m, we consider the acceleration equation:

( v \frac{dv}{dx} = \left( 1 - \frac{v^2}{3200} \right) )

We solve this by separating variables:

$\int_0^v \frac{3200}{3200 - v^2} dv = \int_0^{1500} dx$

After integration, we get:

$-1600\ln(3200 - v^2) = \left[ x \right]_0^{1500}$ ,

This simplifies to:

$1600\ln(3200 - v^2) = -1500$

Thus,

$3200 - v^2 = 3200 e^{-\frac{1500}{1600}}$

Solving for ( v ) yields:

$v \approx 441.2 \, m/s$

Step 3

Calculate correct to two decimal places the maximum speed of the car.

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Answer

To find the maximum speed of the car, we set the acceleration to zero:

$1 - \frac{v^2}{3200} = 0$

Solving for ( v ) gives:

$v^2 = 3200 \Rightarrow v = \sqrt{3200} \approx 56.57 \, m/s$

10. (a) Solve the differential equation \( \frac{dy}{dx} = y^3 \sin x \) given that \( y = 1 \) when \( x = \frac{\pi}{2} \) - Leaving Cert Applied Maths - Question 10 - 2007

Worked Solution & Example Answer:10. (a) Solve the differential equation \( \frac{dy}{dx} = y^3 \sin x \) given that \( y = 1 \) when \( x = \frac{\pi}{2} \) - Leaving Cert Applied Maths - Question 10 - 2007

Solve the differential equation \( \frac{dy}{dx} = y^3 \sin x \)

Calculate correct to two decimal places the speed of the car when it has travelled 1500 m from rest

Calculate correct to two decimal places the maximum speed of the car.

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