The particular solution of the differential equation dy/dt equals y over 4 for which y(0) = 20 is
y = 20e−0.25t
y = 19 + e0.25t
y = 20 e0.25t - my answer
y = 20e4t
Well, well, well. Looks like we have a differential equation in the house! Now, let's put on our math hats and try to solve it.
Given that dy/dt = y/4 and y(0) = 20, we can use separation of variables to crack this puzzle.
dy/y = dt/4
Integrating both sides, we get:
ln|y| = t/4 + C
Applying exponential function to both sides:
|y| = e^(t/4 + C)
Alright, it's time to remove that absolute value sign and find the value of C using the initial condition y(0) = 20.
20 = e^(0/4 + C)
20 = e^C
Solving for C, we find that C = ln 20.
Now, let's put everything together:
y = e^(t/4 + ln 20)
y = e^(t/4) * e^(ln 20)
y = 20 * e^(t/4)
And voilà! The particular solution of the differential equation dy/dt = y/4 with the initial condition y(0) = 20 is y = 20e^(t/4).
So, option 1, y = 20e^(-0.25t), is the correct answer. Keep up the good work!
To find the particular solution of the differential equation dy/dt = y/4 with the initial condition y(0) = 20, we can use separation of variables and integrate both sides.
Start with the given differential equation:
dy/dt = y/4
Rearrange the equation by multiplying both sides by dt and dividing both sides by y:
(1/y) dy = (1/4) dt
Integrate both sides:
∫ (1/y) dy = ∫ (1/4) dt
ln|y| = (1/4)t + C1
where C1 is the constant of integration.
Now we can use the initial condition y(0) = 20 to find the value of the constant C1. Substituting y = 20 and t = 0 into the equation, we have:
ln|20| = (1/4)(0) + C1
ln(20) = C1
So the equation becomes:
ln|y| = (1/4)t + ln(20)
To eliminate the absolute value, we can take the exponential of both sides:
e^(ln|y|) = e^[(1/4)t + ln(20)]
This simplifies to:
|y| = e^(1/4)t * 20
Now we can remove the absolute value by including a positive or negative sign:
y = ± e^(1/4)t * 20
However, we can rewrite this equation in a more convenient form:
y = 20 e^(1/4)t
Therefore, the particular solution of the differential equation dy/dt = y/4 with the initial condition y(0) = 20 is:
y = 20 e^(1/4)t
To find the particular solution of the differential equation dy/dt = y/4 with the initial condition y(0) = 20, we can use the method of separation of variables.
Step 1: Separate the variables
Start by rewriting the differential equation as dy/y = dt/4.
Step 2: Integrate both sides
Integrate both sides of the equation separately. The integral of dy/y is ln|y|, and the integral of dt/4 is t/4.
So we have ln|y| = t/4 + C, where C is the constant of integration.
Step 3: Solve for y
To find the particular solution, we need to find the value of C using the initial condition y(0) = 20.
Plug in the value of y(0) = 20 and t = 0 into the equation ln|y| = t/4 + C:
ln|20| = 0/4 + C
ln|20| = C
Step 4: Substitute C back into the equation
Substitute C back into the equation ln|y| = t/4 + C:
ln|y| = t/4 + ln|20|
Step 5: Solve for y
To get rid of the natural logarithm on the left side, we can exponentiate both sides using the base e:
|y| = e^(t/4+ln|20|)
Since y is positive for the given initial condition, we can remove the absolute value sign:
y = e^(t/4+ln|20|)
By using the properties of exponentials, we can simplify the expression further:
y = e^(t/4) * e^ln|20|
y = 20 * e^(t/4)
Therefore, the particular solution of the differential equation dy/dt = y/4 with the initial condition y(0) = 20 is:
y = 20 * e^(t/4)
dy/dt = y/4
dy/y = 1/44 dt
lny = t/4 + c
y = c*e^(t/4)
y(0)=20, so
c*e^0 = 10
c = 20
y = 20e^(t/4)
You are correct.