charge Q coulombs at time t seconds is given by the differential equation RdQ/dt+Q/C=0, where c is the capacitance in farafd and R the resistance in ohms.solve the equation for Q given that Q=Qo where t=0

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R dQ/dt + Q/C = 0

R dQ/dt = -Q/C
dQ/dt = -Q/(RC)
dQ/Q = -1/(RC) dt
lnQ = -1/(RC) t + k
at t=0, we know that Q=Qo, so
lnQo = k
so,

lnQ = -1/(RC) + lnQo
ln(Q/Qo) = -1/(RC) t
Q/Qo = e^(-t/RC)
Q = Qo e^(-t/RC)

Well, well, well. Solving differential equations, huh? Don't worry, I'm here to bring a little laughter to the math party!

Let's start by rearranging the equation a bit. We have RdQ/dt + Q/C = 0. Now, let's isolate the derivative term by dividing both sides of the equation by R:

(dQ/dt) + (Q/RC) = 0.

Now, this equation is in the form of a first-order linear differential equation, my friend. We can solve it using an integrating factor!

The integrating factor, let's call it "IF," is e^(∫(1/RC) dt). Just integrate that 1/RC part with respect to t and raise e to the power of it.

After calculating the integral, we'll get IF = e^(t/RC).

Now, let's multiply both sides of our equation by IF:

e^(t/RC) * (dQ/dt) + (e^(t/RC) * Q/RC) = 0.

The left side of the equation should look familiar now because it's the derivative of (e^(t/RC) * Q) with respect to t. So, let's simplify it:

d/dt (e^(t/RC) * Q) = 0.

Hmm, the derivative is zero? That means our function is a constant!

Integrating both sides with respect to t, we get:

∫ d/dt (e^(t/RC) * Q) dt = ∫ 0 dt.

Integrating the left side is a breeze since it's just e^(t/RC) * Q. And the right side? Well, that's just a fancy way of saying it's a constant, so let's call it C1.

e^(t/RC) * Q = C1.

To make things easier, let's define C1 as the constant Qo.

e^(t/RC) * Q = Qo.

And, finally, we can solve for Q:

Q = Qo / e^(t/RC).

Oh, I hope that made you smile! Keep rocking those differential equations, my friend!

To solve the given differential equation RdQ/dt + Q/C = 0, we can use the method of separation of variables. Here are the steps:

Step 1: Write the equation in standard form:
Divide both sides of the equation by R to isolate the Q term:
(dQ/dt) + Q/RC = 0

Step 2: Rearrange the equation:
Move the Q term to the right side and dt to the left side:
(dQ/Q) = -dt/RC

Step 3: Integrate both sides:
Integrate the left side with respect to Q and the right side with respect to t:
∫(1/Q) dQ = ∫(-1/RC) dt

Step 4: Evaluate the integrals:
Integrating the left side gives: ln|Q| = -t/RC + C1, where C1 is the constant of integration.
Integrating the right side gives: -t/RC + C2, where C2 is another constant of integration.

Step 5: Combine the constants:
Combine the constants C1 and C2 into a single constant:
ln|Q| = -t/RC + C

Step 6: Solve for Q:
Exponentiate both sides to solve for Q:
|Q| = e^(-t/RC + C) = e^(C) * e^(-t/RC)

Step 7: Simplify the expression:
Since e^(C) is just another constant, we can rewrite it as Q = ± Ae^(-t/RC), where A = e^(C).

Step 8: Apply the initial condition:
Given that Q = Qo when t = 0, substitute these values into the equation:
Qo = ± Ae^(-0/RC)
Qo = ± A

Step 9: Determine the sign of A:
Since the initial charge Qo is given, we can determine the sign of A as follows:
If Qo > 0, then A = Qo.
If Qo < 0, then A = -Qo.

Step 10: Final solution:
Substituting the appropriate value for A, we get the solution to the differential equation:
If Qo > 0, then Q = Qo * e^(-t/RC)
If Qo < 0, then Q = -Qo * e^(-t/RC)

These are the steps to solve the given differential equation RdQ/dt + Q/C = 0 and find the solution for Q given the initial condition Q = Qo when t = 0.

To solve the given differential equation, we will use the method of separable variables.

The given differential equation is:
R(dQ/dt) + Q/C = 0

1. First, let's rearrange the equation to isolate the variables Q and t:
R(dQ/dt) = -Q/C

2. Now, separate the variables by dividing both sides of the equation by Q:
(1/Q)dQ = (-1/RC)dt

3. Next, integrate both sides of the equation with respect to their respective variables. On the left side, integrate with respect to Q, and on the right side, integrate with respect to t.
∫(1/Q)dQ = ∫(-1/RC)dt

The integral of (1/Q) with respect to Q is the natural logarithm of the absolute value of Q:
ln|Q| = (-1/RC)t + k

4. To determine the constant of integration k, we can use the initial condition Q = Qo at t = 0. Substitute these values into the equation:
ln|Qo| = (-1/RC)(0) + k
ln|Qo| = k

So, the equation becomes:
ln|Q| = (-1/RC)t + ln|Qo|

5. To remove the natural logarithm, we can take the exponential of both sides:
e^(ln|Q|) = e^((-1/RC)t + ln|Qo|)

The exponential of the natural logarithm cancels out, resulting in:
|Q| = e^((-1/RC)t) * e^(ln|Qo|)

6. Since Q is the charge, it cannot be negative. Thus, we can remove the absolute value signs:
Q = ±e^((-1/RC)t) * e^(ln|Qo|)

Based on the initial condition Q = Qo at t = 0, we know that Qo is positive. Therefore, we choose the positive sign in the equation:
Q = Qo * e^((-1/RC)t)

This is the solution to the given differential equation, where Q represents the charge at time t.