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faltersainse [42]
3 years ago
7

Solve the seperable equation: dx/dt=3xt^2

Mathematics
1 answer:
andrew-mc [135]3 years ago
5 0
\displaystyle
\dfrac{dx}{dt}=3xt^2\\
\dfrac{dx}{x}=3t^2 \, dt\\
\int \dfrac{dx}{x}=\int 3t^2 \, dt\\
\ln x=3\dfrac{t^3}{3}+C\\
\ln x=t^3+C\\
x=e^{t^3+C}\\
x=e^{t^3}\cdot e^C\\
\boxed{x=Ce^{t^3}}
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Answer:

constant of variation = 3

Step-by-step explanation:

we know that b varies jointly with c and d

so:

b∝c∝d

and b varies inversely with e, so

b∝\frac{1}{e}

and i will call the constant of variation k, this way we can make an equation for b in the following form:

b=k\frac{cd}{e}

this satisfy that b varies jointly with c and d (if b increases, c and d also increase) and inversely with e (if b increases, e decreases)

we know that when b is 18, c is 4, d is 9, and e is 6:

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substituting this in our equation for b:

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and we solve operations and clear for the constant of variation k:

18=k\frac{36}{6}\\ 18=6k\\\frac{18}{6}=k\\ 3=k

the constant of variation is 3.

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