The particular solution of the differential equation dy/dt=y/4 for which y(0) = 20 is
y = 20e^-0.25t
y = 19 + e^0.25t
y = 20 e^0.25t
y = 20^e4t
Looks like (c) to me
To find the particular solution of the given differential equation, we can begin by separating the variables and integrating both sides.
The given differential equation is: dy/dt = y/4.
Separating the variables, we can move all the y terms to one side and all the t terms to the other side:
(1/y) dy = (1/4) dt.
Now, we can integrate both sides of the equation.
∫ (1/y) dy = ∫ (1/4) dt.
Integrating the left side with respect to y gives us: ln|y| = (1/4)t + C1, where C1 is the constant of integration.
To isolate y, we can apply the exponential function to both sides to get rid of the natural logarithm:
e^(ln|y|) = e^((1/4)t + C1).
Using the property of exponential and logarithmic functions, we simplify this equation to:
|y| = e^((1/4)t) * e^(C1).
Now, we can rewrite the equation using the absolute value notation:
y = ± e^((1/4)t) * e^(C1).
Next, we incorporate the initial condition y(0) = 20. This gives us:
20 = ± e^((1/4)*0) * e^(C1).
Simplifying further, we have:
20 = ± e^0 * e^(C1) = ± 1 * e^(C1) = ± e^(C1).
Since e^(C1) is a positive constant according to our initial condition, we can rewrite the equation as:
20 = e^(C1).
Now, we substitute this value of e^(C1) back into our expression for y:
y = ± e^((1/4)t) * 20.
Finally, we can simplify further by dropping the ± sign, as it is not needed due to the initial condition being given. Therefore, the particular solution of the differential equation dy/dt = y/4 for y(0) = 20 is:
y = 20e^((1/4)t).