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Parents Pricing Home Leaving Cert Applied Maths Differential Equations If
$$x^3 y'\frac{dy}{dx} + y'\frac{dy}{dx} = 1$$
and $y = 0$ when $x = 0$, find the value of $x$ when $y = \frac{\pi}{2}$
If
$$x^3 y'\frac{dy}{dx} + y'\frac{dy}{dx} = 1$$
and $y = 0$ when $x = 0$, find the value of $x$ when $y = \frac{\pi}{2}$ - Leaving Cert Applied Maths - Question 10 - 2008 Question 10
View full question If
$$x^3 y'\frac{dy}{dx} + y'\frac{dy}{dx} = 1$$
and $y = 0$ when $x = 0$, find the value of $x$ when $y = \frac{\pi}{2}$.
(b) A train of mass 200 tonnes moves a... show full transcript
View marking scheme Worked Solution & Example Answer:If
$$x^3 y'\frac{dy}{dx} + y'\frac{dy}{dx} = 1$$
and $y = 0$ when $x = 0$, find the value of $x$ when $y = \frac{\pi}{2}$ - Leaving Cert Applied Maths - Question 10 - 2008
If $x^3 y'\frac{dy}{dx} + y'\frac{dy}{dx} = 1$ Only available for registered users.
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To solve this differential equation, we can separate the variables and integrate:
Rearranging gives us:
d y d x ( x 3 + 1 ) = 1 \frac{dy}{dx}(x^3 + 1) = 1 d x d y ( x 3 + 1 ) = 1 d y = 1 x 3 + 1 d x dy = \frac{1}{x^3 + 1} dx d y = x 3 + 1 1 d x
Integrate both sides:
∫ d y = ∫ 1 x 3 + 1 d x \int dy = \int \frac{1}{x^3 + 1} dx ∫ d y = ∫ x 3 + 1 1 d x
The left side integrates to:
y = 1 2 ln ( 1 + x 3 ) + C y = \frac{1}{2} \ln(1 + x^3) + C y = 2 1 ln ( 1 + x 3 ) + C
Given that y = 0 y = 0 y = 0 x = 0 x = 0 x = 0 C C C 0 = 0 + C ⇒ C = 0 0 = 0 + C \Rightarrow C = 0 0 = 0 + C ⇒ C = 0
Therefore, the equation simplifies to:
y = 1 2 ln ( 1 + x 3 ) y = \frac{1}{2} \ln(1 + x^3) y = 2 1 ln ( 1 + x 3 )
Setting y = π 2 y = \frac{\pi}{2} y = 2 π x x x
Find the value of P. Only available for registered users.
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Start by using Newton's second law, applying it to this train system:
T = 1000 P v T = \frac{1000P}{v} T = v 1000 P
The force equals to the mass times acceleration, giving us:
F = 200000 ⋅ ( 8000 − v 3 500 v ) F = 200000 \cdot (\frac{8000 - v^3}{500v}) F = 200000 ⋅ ( 500 v 8000 − v 3 )
Therefore:
\n1000 P v − 400 v 2 = 200000 ⋅ 8000 − v 3 500 v \frac{1000P}{v} - 400v^2 = 200000 \cdot \frac{8000 - v^3}{500v} v 1000 P − 400 v 2 = 200000 ⋅ 500 v 8000 − v 3
Rearranging provides:
1000 P = 320000 − 400 v 3 1000P = 320000 - 400v^3 1000 P = 320000 − 400 v 3
Solving for P P P P = 3200. P = 3200. P = 3200.
The train travels a distance 69.07 m while its speed increases from 10 m/s to $v_1$ m/s. Only available for registered users.
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The work done over the distance relates to the change in kinetic energy by:
∫ 10 v 1 500 v 8000 − v 3 d s = ∫ 0 69.07 d x \int_{10}^{v_1} \frac{500v}{8000 - v^3} ds = \int_0^{69.07} dx ∫ 10 v 1 8000 − v 3 500 v d s = ∫ 0 69.07 d x
Consequently, we get:
∫ 10 v 1 500 8000 − v 3 d v = 69.07 \int_{10}^{v_1} \frac{500}{8000 - v^3} dv = 69.07 ∫ 10 v 1 8000 − v 3 500 d v = 69.07
Solving the integrals:
[ − 500 3 ln ( 8000 − v 3 ) ] 10 v 1 = 69.07 \left[ -\frac{500}{3} \ln(8000 - v^3) \right]_{10}^{v_1} = 69.07 [ − 3 500 ln ( 8000 − v 3 ) ] 10 v 1 = 69.07
Evaluating gives:
− 500 3 ln ( 8000 − v 1 3 ) + 500 3 ln ( 8000 − 1000 ) = 69.07 -\frac{500}{3} \ln(8000 - v_1^3) + \frac{500}{3} \ln(8000 - 1000) = 69.07 − 3 500 ln ( 8000 − v 1 3 ) + 3 500 ln ( 8000 − 1000 ) = 69.07
Thus we find:
ln ( 8000 − v 1 3 ) = 69.07 ⋅ 3 500 + ln ( 7000 ) \ln(8000 - v_1^3) = \frac{69.07 \cdot 3}{500} + \ln(7000) ln ( 8000 − v 1 3 ) = 500 69.07 ⋅ 3 + ln ( 7000 )
Evaluating results in:
v 1 = 15.0 m / s . v_1 = 15.0 m/s. v 1 = 15.0 m / s .
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