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muminat
2 years ago
7

Giải phương trình y"= lnx

Mathematics
1 answer:
kipiarov [429]2 years ago
3 0

Integrate both sides with respect to <em>x</em> :

y''(x) = \ln(x) \implies \displaystyle \int y''(x)\,\mathrm dx = \int \ln(x)\,\mathrm dx

On the right side, integrate by parts with

f = \ln(x) \implies \mathrm df = \dfrac{\mathrm dx}x \\\\ \mathrm dg = \mathrm dx \implies g = x

\implies \displaystyle \int\ln(x)\,\mathrm dx = fg - \int g\,\mathrm df \\\\ \int\ln(x)\,\mathrm dx= x\ln(x) - \int \mathrm dx \\\\ \int\ln(x)\,\mathrm dx= x\ln(x) - x + C_1

Then

\displaystyle y'(x) = x\ln(x) - x + C_1

Integrate both sides with respect to <em>x</em> by parts again, this time with

f = \ln(x) \implies \mathrm df = \dfrac{\mathrm dx}x \\\\ \mathrm dg = x\,\mathrm dx \implies g = \dfrac{x^2}2

\implies \displaystyle \int x\ln(x)\,\mathrm dx = fg-\int g\,\mathrm df \\\\ \int x\ln(x)\,\mathrm dx = \dfrac{x^2\ln(x)}2 - \int\frac x2\,\mathrm dx \\\\ \int x\ln(x)\,\mathrm dx = \dfrac{x^2\ln(x)}2-\dfrac{x^2}4 + C_2

Then

y(x) = \dfrac{x^2\ln(x)}2 - \dfrac{x^2}4 - \dfrac{x^2}2 + C_1x + C_2 \\\\ \boxed{y(x) = \dfrac{(2\ln(x)-3)x^2}4 + C_1x + C_2}

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I need help on this question please
Fynjy0 [20]

Answer: NEITHER

Step-by-step explanation:

From the equation of lines given above, to know if relationships exist between the two lines, we must  first determine that  from their gradients or slopes.

From first equation

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      y = 2x - 5

Therefore, m₁ = 2

From the second equation

3x - y = 5

       y = 3x - 5

Therefore m₂ = 3

Recall, For the two lines to be parallel, m₁ = m₂ , and for the two line to be perpendicular, m₁m₂ = -1

since none fulfilled the conditions stated above, the answer then is

NEITHER

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3 years ago
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Answer:

The change in temperature per minute for the sample, dT/dt is 71.\overline {6} °C/min

Step-by-step explanation:

The given parameters of the question are;

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Given that both dQ/dT and dQ/dt are known, we have;

\dfrac{dQ}{dT} = 0.18 \, (kcal/ ^{\circ} C)

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Therefore, we get;

\dfrac{\dfrac{dQ}{dt} }{\dfrac{dQ}{dT} } = {\dfrac{dQ}{dt} } \times \dfrac{dT}{dQ} = \dfrac{dT}{dt}

\dfrac{dT}{dt} = \dfrac{\dfrac{dQ}{dt} }{\dfrac{dQ}{dT} } = \dfrac{12.9 \, kcal / min }{0.18 \, kcal/ ^{\circ} C }   = 71.\overline 6 \, ^{\circ } C/min

For the sample, we have the change in temperature per minute, dT/dt, presented as follows;

\dfrac{dT}{dt}  = 71.\overline 6 \, ^{\circ } C/min

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Step-by-step explanation:

1.5 : 2.5

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15/25

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