Elementary differential equations

[kristo]

Homework Statement

I can't solve the following problems:
38) Solve y' =y + e^t from yo =0 by Method 2, where the
deposit e^T at time T is multiplied by e^t-T. The total output
at time t is y(t) = (Integral)(e^T * e^t-TdT) (The integral goes from 0 to t). Substitute back to
check y' =y + e^t.
39) Rewrite y' =y + e^t as y' - y = e^t. Multiplying by e^-t, the
left side is the derivative of _________. Integrate both sides
from y₀ = 0 to find y(t).
40) Solve y' = -y + 1 from y₀ =0 by rewriting as y' + y = 1,
multiplying by e^t, and integrating both sides.
41) Solve y' =y + t from y₀ = 0 by assuming y = A*e^t + B*t + C.

The Method 2 referred to in problem 38 is explained thus:
The input y₀ produces the output y₀e^ct. After t
years any deposit is multiplied by e^ct. That also applies to deposits made after the account is opened. If the deposit enters at time T the growing time is only t -T
Therefore the multiplying factor is only e^c(t-T).This growth factor applies to the small deposit (amount s*dT) made between time T and T + dT.
Now add up all outputs at time t. The output from y₀ is y₀e^ct. The small deposit
s*dT near time T grows to e^c(t-T)*s*dT. The total is an integral:

y(t) = y₀e^ct + (Integral)(e^c(t-T)*s*dT) (Integral from 0 to t)


Sorry about the messy equations, but I couldn't get the Integral sign to work, it messed up the equations every time I tried.

Homework Equations

The Attempt at a Solution

I have no idea where to start. I can't figure out how to apply the Method 2 in problem 38 and in others I don't know how to integrate a function of two variables (they all depend on both y and t, right?)
Some hints, please.

 

[mighty2000]

I understand that these problems may seem difficult, but don't worry, there are a few key steps that can help you solve them. Let's break them down one by one.

38) To solve this problem, we can use the Method 2 described in the problem statement. This method tells us that the input y0 produces the output y0ect, and any deposit made after time T is multiplied by ect-T. So, to find the total output at time t, we need to add up all the outputs from y0 to t. This can be done by using an integral, which represents the sum of all the small deposits made between T and t. The integral will go from T to t, since we are only considering deposits made after T. So the total output at time t will be given by the integral (Integral)(eT * et-TdT) from T to t. This is the equation given in the problem statement. To check that this is the correct solution, we can substitute it back into the original equation y' = y + et and see if it holds true.

39) In this problem, we are asked to rewrite the equation y' = y + et in a different form. We can do this by subtracting y from both sides, which gives us y' - y = et. Now, if we multiply both sides by e-t, we get (e-t)(y' - y) = et * e-t, which simplifies to y' * e-t - y * e-t = 1. Notice that the left side is now the derivative of y * e-t with respect to t. So if we integrate both sides with respect to t, we get y * e-t = t + C, where C is a constant of integration. Solving for y, we get y = (t + C) * e-t. To find the value of C, we can substitute in the initial condition y0 = 0, which gives us C = -1. So the final solution is y = (t - 1) * e-t.

40) This problem is similar to problem 39, but instead of rewriting the equation, we are asked to solve it by rewriting it in a different form. We can do this by adding y to both sides, which gives us y' + y = 1. Now, if we multiply both sides by et and integrate with respect to t