Show that 𝑞(𝑡) = 𝑞0(𝑒−𝑡/𝜏) is a solution to this differential equation (𝑅 (𝑑𝑞/𝑑𝑡) = −𝑞/𝐶) if 𝜏 = 𝑅𝐶.

Question

Show that 𝑞(𝑡) = 𝑞0(𝑒−𝑡/𝜏) is a solution to this differential equation (𝑅 (𝑑𝑞/𝑑𝑡) = −𝑞/𝐶) if 𝜏 = 𝑅𝐶.

Answer 1

To show that 𝑞(𝑡) = 𝑞0(𝑒^(-𝑡/𝜏)) is a solution to the differential equation 𝑅 (𝑑𝑞/𝑑𝑡) = -𝑞/𝐶, we need to substitute 𝑞(𝑡) into the differential equation and verify that it satisfies the equation.

Given 𝜏 = 𝑅𝐶, we have:

𝑞(𝑡) = 𝑞0*(𝑒^(-𝑡/𝜏))

Now let's calculate the derivative of 𝑞(𝑡) with respect to 𝑡:

𝑑𝑞/𝑑𝑡 = 𝑞0 * (-1/𝜏) * (𝑒^(-𝑡/𝜏)) [using chain rule]

Next, let's substitute 𝑞(𝑡) and its derivative into the differential equation:

𝑅 * (𝑑𝑞/𝑑𝑡) = -𝑞/𝐶

𝑅 * (𝑞0 * (-1/𝜏) * (𝑒^(-𝑡/𝜏))) = -𝑞0*(𝑒^(-𝑡/𝜏))/𝐶

Multiplying both sides by -𝜏/𝑞0, we get:

𝑅 * (-1/𝜏) * (𝑒^(-𝑡/𝜏)) = -(𝑒^(-𝑡/𝜏))/𝐶

Dividing both sides by -1/𝐶, we get:

𝑅 * (𝑒^(-𝑡/𝜏)) = 𝑒^(-𝑡/𝜏)/𝐶

Multiplying both sides by 𝜏, we get:

𝜏 * 𝑅 * (𝑒^(-𝑡/𝜏)) = (𝑡/𝜏)/𝐶

Since we know that 𝜏 = 𝑅𝐶, we can substitute it into the equation:

𝑅 * 𝑅𝐶 * (𝑒^(-𝑡/(𝑅𝐶))) = (𝑡/(𝑅𝐶))/𝐶

Simplifying both sides, we get:

𝑅^2 * 𝑒^(-𝑡/(𝑅𝐶)) = (𝑡/𝐶^2)

Now notice that 𝑡/𝐶^2 is just a constant, so let's denote it as 𝐴:

𝑅^2 * 𝑒^(-𝑡/(𝑅𝐶)) = 𝐴

Taking the natural logarithm of both sides to solve for 𝑒^(-𝑡/(𝑅𝐶)), we get:

-𝑡/(𝑅𝐶) = ln(𝐴)/𝑅^2

Multiplying both sides by -1 and 𝑅𝐶, we get:

𝑡 = -𝑅𝐶 * ln(𝐴)/𝑅^2

Simplifying the right side, we have:

𝑡 = -(𝐵/𝑅) * (1/𝑅) * ln(𝐴)

𝑡 = -(𝐵/𝑅^2) * ln(𝐴)

Noting that -(𝐵/𝑅^2) is just a constant, let's denote it as 𝐶1:

𝑡 = 𝐶1 * ln(𝐴)

Therefore, 𝑡 is a linear function of ln(𝐴), which implies that 𝑒^(-𝑡/(𝑅𝐶)) = 𝐴, where 𝐴 is a constant.

This confirms that 𝑞(𝑡) = 𝑞0*(𝑒^(-𝑡/𝜏)) satisfies the differential equation 𝑅 *(𝑑𝑞/𝑑𝑡) = -𝑞/𝐶.

Answer 2

To show that 𝑞(𝑡) = 𝑞0 * (𝑒^(-𝑡/𝜏)) is a solution to the differential equation 𝑅 * (𝑑𝑞/𝑑𝑡) = -𝑞/𝐶, we need to substitute 𝑞(𝑡) and its derivative into the equation and show that it satisfies the equation.

First, let's find the derivative of 𝑞(𝑡) with respect to 𝑡:

𝑞(𝑡) = 𝑞0 * (𝑒^(-𝑡/𝜏))
Taking the derivative of both sides with respect to 𝑡:
𝑑𝑞/𝑑𝑡 = 𝑞0 * (-1/𝜏) * (𝑒^(-𝑡/𝜏))

Now, substitute 𝑞(𝑡) and its derivative into the differential equation 𝑅 * (𝑑𝑞/𝑑𝑡) = -𝑞/𝐶:

𝑅 * (𝑑𝑞/𝑑𝑡) = -𝑞/𝐶
𝑅 * (𝑞0 * (-1/𝜏) * (𝑒^(-𝑡/𝜏))) = -𝑞/𝐶
𝑅 * (-𝑞0/𝜏) * (𝑒^(-𝑡/𝜏)) = -𝑞/𝐶

Next, we can simplify the left-hand side of the equation:

𝑅 * (-𝑞0/𝜏) * (𝑒^(-𝑡/𝜏)) = (-𝑅𝑞0/𝜏) * (𝑒^(-𝑡/𝜏))

By observing that 𝑅𝑞0/𝜏 = -𝑞/𝐶 (since 𝜏 = 𝑅𝐶), we can rewrite the equation as:

(-𝑞/𝐶) * (𝑒^(-𝑡/𝜏)) = -𝑞/𝐶

As we can see, both sides of the equation are equal, thereby confirming that 𝑞(𝑡) = 𝑞0 * (𝑒^(-𝑡/𝜏)) is a solution to the given differential equation 𝑅 * (𝑑𝑞/𝑑𝑡) = -𝑞/𝐶.

Show that 𝑞(𝑡) = 𝑞0*(𝑒−𝑡/𝜏) is a solution to this differential equation (𝑅 *(𝑑𝑞/𝑑𝑡) = −𝑞/𝐶) if 𝜏 = 𝑅𝐶.

To show that 𝑞(𝑡) = 𝑞0*(𝑒^(-𝑡/𝜏)) is a solution to the differential equation 𝑅 *(𝑑𝑞/𝑑𝑡) = -𝑞/𝐶, we need to substitute 𝑞(𝑡) into the differential equation and verify that it satisfies the equation.

To show that 𝑞(𝑡) = 𝑞0 * (𝑒^(-𝑡/𝜏)) is a solution to the differential equation 𝑅 * (𝑑𝑞/𝑑𝑡) = -𝑞/𝐶, we need to substitute 𝑞(𝑡) and its derivative into the equation and show that it satisfies the equation.

Show that 𝑞(𝑡) = 𝑞0(𝑒−𝑡/𝜏) is a solution to this differential equation (𝑅 (𝑑𝑞/𝑑𝑡) = −𝑞/𝐶) if 𝜏 = 𝑅𝐶.

To show that 𝑞(𝑡) = 𝑞0(𝑒^(-𝑡/𝜏)) is a solution to the differential equation 𝑅 (𝑑𝑞/𝑑𝑡) = -𝑞/𝐶, we need to substitute 𝑞(𝑡) into the differential equation and verify that it satisfies the equation.