Verify if the given equations are solutions to the differential equation.
dy/dt=k(y-a)
i) y=1-ae^(kt)
ii) y=a+Ce^(kt)
dy/dt=k(y-a) ... (1)
i) y=1-ae^(kt) (particular solution)
differentiate y w.r.t. t:
d(1-ae^(kt))/dt
= k(e^(kt)+a)-ka
= ky-ka
= k(y-a)
= right-hand side, QED
ii) y=a+Ce^(kt)
differentiate with respect to t:
dy/dt
=Cke^(kt)
=k(a+Ce^(kt))-ka
=ky-ka
=k(y-a)
=right-hand side QED
To verify if the given equations are solutions to the differential equation dy/dt = k(y-a), we need to substitute them into the differential equation and see if they satisfy it.
i) y = 1 - ae^(kt)
To check if this equation is a solution, we need to find its derivative:
dy/dt = d/dt (1 - ae^(kt))
= 0 - a(ke^(kt))
= -ak(e^(kt))
Now substitute this derivative and the given y value into the differential equation:
-ak(e^(kt)) = k(1 - ae^(kt) - a)
Simplifying this equation:
-ak(e^(kt)) = k - kae^(kt) - ak
Multiply through by -1:
ak(e^(kt)) = -k + kae^(kt) + ak
ak(e^(kt)) + k + kae^(kt) + ak = 0
ak(e^(kt) + 1 + ae^(kt) + 1) = 0
ak(e^(kt) + ae^(kt) + 2) = 0
Since k, a, and e^(kt) are all positive, this equation cannot be true for all values of t. Therefore, the equation y = 1 - ae^(kt) is NOT a solution to the given differential equation.
ii) y = a + Ce^(kt)
To check if this equation is a solution, we need to find its derivative:
dy/dt = d/dt (a + Ce^(kt))
= 0 + C(ke^(kt))
= Cke^(kt)
Now substitute this derivative and the given y value into the differential equation:
Cke^(kt) = k(a + Ce^(kt) - a)
Simplifying this equation:
Cke^(kt) = kCe^(kt)
Dividing through by k and e^(kt):
C = C
This equation is true for all values of C. Therefore, the equation y = a + Ce^(kt) is a solution to the given differential equation.
In summary:
i) y = 1 - ae^(kt) is NOT a solution to the differential equation.
ii) y = a + Ce^(kt) is a solution to the differential equation.
To verify if the given equations are solutions to the differential equation dy/dt = k(y - a), we need to substitute the equations into the differential equation and check whether they satisfy it.
i) y = 1 - ae^(kt)
First, let's find dy/dt:
dy/dt = d/dt(1 - ae^(kt))
Using the chain rule, we have:
dy/dt = 0 - ake^(kt) = -ak * e^(kt)
Now, let's substitute y and dy/dt into the differential equation:
-ak * e^(kt) = k(1 - ae^(kt) - a)
Distribute k on the right side:
-ak * e^(kt) = k - kae^(kt) - ka
Combine like terms:
-ak * e^(kt) = k - ka - kae^(kt)
Rearranging the equation:
-ak * e^(kt) + ka + kae^(kt) - k = 0
We can factor out -ak and simplify:
e^(kt)(-ak + ka) + k - k = 0
Since (-ak + ka) = 0, the equation simplifies to:
0e^(kt) = 0
This equation holds true for all values of t, so the equation y = 1 - ae^(kt) is a solution to the differential equation dy/dt = k(y - a).
ii) y = a + Ce^(kt)
First, let's find dy/dt:
dy/dt = d/dt(a + Ce^(kt))
Using the chain rule, we have:
dy/dt = 0 + Ck * e^(kt) = Ck * e^(kt)
Now, let's substitute y and dy/dt into the differential equation:
Ck * e^(kt) = k(a + Ce^(kt) - a)
Distribute k on the right side:
Ck * e^(kt) = Ck * e^(kt)
The equation simplifies to itself. This means that the equation y = a + Ce^(kt) is also a solution to the differential equation dy/dt = k(y - a).
In conclusion, both i) y = 1 - ae^(kt) and ii) y = a + Ce^(kt) are solutions to the differential equation dy/dt = k(y - a).