#### Daniel Garla Pismel

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dx/dt = (x(t) + 105)/65

x = (x(t) = 105)/65 * t

I am not sure if I can do this because x(t) is a function of time that I don't know. Can anyone help me out and explain it to me?

- Thread starter Daniel Garla Pismel
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dx/dt = (x(t) + 105)/65

x = (x(t) = 105)/65 * t

I am not sure if I can do this because x(t) is a function of time that I don't know. Can anyone help me out and explain it to me?

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Will it give the same answer as the way I did gave? If no, I didn't get it\(\displaystyle \frac{dx}{dt} = \frac{x+105}{65}\)

separate variables ...

\(\displaystyle \frac{dx}{x+105} = \frac{dt}{65}\)

now integrate both sides ...

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Willitgive the same answer as the way I did gave? If no, I didn't get it

Your answer (OP) was absolutely misguided (not even wrong!)

Integrate the "expressions" derived in response #2 - tell us what you found.

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What's the point of writing like this? The only thing you achieved was making me feel bad

And the answer I've got now is x = 1,613t/(1-t/65). Is this right?

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Then using your method \(\displaystyle \int xdt = xt +c = x\sqrt{x} + c= x^{3/2}+ c\)

Since x=t^2, those two parallel lines between x and t^2 tells us we can replace t^2 with x and x with t^2 whenever we like.

So \(\displaystyle \int xdt = \int t^2 dt = \dfrac{t^3}{3} + c \neq x^{3/2}+ c = (t^2)^{3/2} + c = t^3\)

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The point was to make you review your response - and find your "mistake".

What's the point of writing like this?The only thing you achieved was making me feel bad

And the answer I've got now is x = 1,613t/(1-t/65). Is this right?

Now you found:

x = 1,613t/(1-t/65)

Differentiate "your answer" and see if it satisfies the given DE. If it does then you

At a first glance

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Thank you, guys, I got it now. Thank you

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from your original post ...Can I integrate the following with respect to t:

dx/dt = (x(t) + 105)/65

x = (x(t) = 105)/65 * t

\(\displaystyle \frac{dx}{dt} = \frac{x+105}{65}\)

separate variables ...

\(\displaystyle \frac{dx}{x+105} = \frac{dt}{65}\)

integrate ...

\(\displaystyle \ln|x+105| = \frac{t}{65} + C_1\)

\(\displaystyle x+105 = e^{\frac{t}{65} + C_1}\)

\(\displaystyle x = C_2e^{\frac{t}{65}} - 105\)