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Kobotan [32]
3 years ago
14

Janice is thinking of two numbers. She says that two times the first number plus the second number is 41. In​ addition, the firs

t number plus three times the second number is 73. Find the numbers
Mathematics
1 answer:
Rudiy273 years ago
7 0
Let one of the number be x and the other be y

<span>two times the first number plus the second number is 41
</span>2x + y = 41

<span>the first number plus three times the second number is 73
x + 3y = 73

</span>2x + y = 41 --------- (1)
x + 3y = 73 --------- (2)

From (1):
2x + y = 41
y = 41 - 2x ------- sub into (2)
x + 3(41-2x) = 73
x + 123 - 6x = 73
x -6x = 73 - 123
-5x = -50
x = 10 -------- sub into (1)
2 (10) + y = 41
20 + y = 41
y = 41 - 20
y = 21

Ans: x = 10, y = 21
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The first equation is linear:

x\dfrac{\mathrm dy}{\mathrm dx}-y=x^2\sin x

Divide through by x^2 to get

\dfrac1x\dfrac{\mathrm dy}{\mathrm dx}-\dfrac1{x^2}y=\sin x

and notice that the left hand side can be consolidated as a derivative of a product. After doing so, you can integrate both sides and solve for y.

\dfrac{\mathrm d}{\mathrm dx}\left[\dfrac1xy\right]=\sin x
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- - -

The second equation is also linear:

x^2y'+x(x+2)y=e^x

Multiply both sides by e^x to get

x^2e^xy'+x(x+2)e^xy=e^{2x}

and recall that (x^2e^x)'=2xe^x+x^2e^x=x(x+2)e^x, so we can write

(x^2e^xy)'=e^{2x}
\implies x^2e^xy=\displaystyle\int e^{2x}\,\mathrm dx=\frac12e^{2x}+C
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- - -

Yet another linear ODE:

\cos x\dfrac{\mathrm dy}{\mathrm dx}+\sin x\,y=1

Divide through by \cos^2x, giving

\dfrac1{\cos x}\dfrac{\mathrm dy}{\mathrm dx}+\dfrac{\sin x}{\cos^2x}y=\dfrac1{\cos^2x}
\sec x\dfrac{\mathrm dy}{\mathrm dx}+\sec x\tan x\,y=\sec^2x
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- - -

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The integrating factor is a function \mu(x) such that

\mu(x)y'(x)+\mu(x)P(x)y(x)=(\mu(x)y(x))'

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