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Could you please solve it?16. = dx dt = 5 dax + 25x = 0, x = 5 at t= 0, and dt2 when t = 7/4. 5

User Amir Keibi
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1 Answer

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Hi! I'd be happy to help you with this differential equation problem. Based on the given information, we have the following:

1. dx/dt = 5(dax) + 25x = 0
2. x(0) = 5
3. Find d^2x/dt^2 when t = 7/4

First, let's rewrite the given differential equation:
dx/dt = -25x/5

Now, separate variables and integrate:

∫(dx/x) = -∫(5 dt)

ln|x| = -5t + C1

Now, exponentiate both sides:

x(t) = e^(-5t + C1)

To find the constant C1, use the initial condition x(0) = 5:

5 = e^(0 + C1)

C1 = ln(5)

So, the solution for x(t) is:

x(t) = e^(-5t + ln(5))

Now we need to find the second derivative, d^2x/dt^2:

dx/dt = -5e^(-5t + ln(5))
d^2x/dt^2 = 25e^(-5t + ln(5))

Finally, find d^2x/dt^2 when t = 7/4:

d^2x/dt^2 (7/4) = 25e^(-5(7/4) + ln(5))

This is the second derivative of x(t) at t = 7/4.

User Star Brilliant
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