dydt=10−0.2y, y(0)=10.



a. Solve this differential equation,

I tried solving it and I keep on getting 10. I dont know how to approach this question

dy/dt = 10−0.2y

5 dy/dt = 50-y
dy/(50-y) = .2 dt
-ln(50-y) = .2t+c
y = ce^-.2t + 50
y(0) = 10, so
10 = c+50
c = -40
y = 50 - 40e^-.2t

Do I plug in 0 for t?

To solve this differential equation dy/dt = 10 - 0.2y, where y(0) = 10, we can use separation of variables method.

Step 1: Separate the variables by moving all the terms involving y to one side and all the terms involving t to the other side:

dy / (10 - 0.2y) = dt

Step 2: Integrate both sides with respect to their respective variables:

∫ (1 / (10 - 0.2y)) dy = ∫ dt

Step 3: Evaluate the integral:

Using the substitution u = 10 - 0.2y, we can rewrite the integral on the left-hand side as:

-5 ∫ (1/u) du

= -5 ln|u| + C

= -5 ln|10 - 0.2y| + C

And the integral on the right-hand side is simply t + C.

Step 4: Combine the results and solve for y:

-5 ln|10 - 0.2y| + C = t + C

-5 ln|10 - 0.2y| = t

ln|10 - 0.2y| = -t/5

Take the exponential of both sides:

10 - 0.2y = e^(-t/5)

Solve for y:

0.2y = 10 - e^(-t/5)

y = (10 - e^(-t/5)) / 0.2

Therefore, the solution to the differential equation is y = (10 - e^(-t/5)) / 0.2, where y(0) = 10.

To solve the given differential equation, you can use the method of separation of variables. Here's how you can approach it:

1. Given differential equation: dy/dt = 10 - 0.2y

2. Rearrange the equation to have all y terms on one side and all t terms on the other side:
dy/(10 - 0.2y) = dt

3. Now, you can integrate both sides of the equation. Integrate the left side with respect to y and the right side with respect to t:
∫(1/(10 - 0.2y))dy = ∫dt

4. Evaluate the integrals:
(1/-0.2)ln|10 - 0.2y| = t + C

5. Simplify the left side:
-5ln|10 - 0.2y| = t + C

6. Rewrite the equation in exponential form in order to isolate y:
ln|10 - 0.2y| = -(t + C)/5

7. Remove the natural logarithm by exponentiating both sides:
|10 - 0.2y| = e^(-(t + C)/5)

8. Eliminate the absolute value by considering the positive and negative cases:
10 - 0.2y = e^(-(t + C)/5) OR 10 - 0.2y = -e^(-(t + C)/5)

9. Solve for y in each case:
Case 1: 10 - 0.2y = e^(-(t + C)/5)
0.2y = 10 - e^(-(t + C)/5)
y = (10 - e^(-(t + C)/5))/0.2

Case 2: 10 - 0.2y = -e^(-(t + C)/5)
0.2y = 10 + e^(-(t + C)/5)
y = (10 + e^(-(t + C)/5))/0.2

10. Finally, to find the specific solution given the initial condition y(0) = 10, substitute t = 0 and y = 10 into the general solution obtained in step 9. Solve for the constant C by plugging in these values:
y(0) = (10 - e^(-0 - C)/5))/0.2 = 10

Now, you can solve for C using algebraic manipulation.