Solve: dA/dt = -10A, A(0) = 5
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I tried getting A by itself:
A = -1/10 * dA/dt
A(0) = 5 = -1/10 * dA/dt
but I'm somewhat sure that's wrong; so integrating I get:
int(dA/dt) = int(-10A)
A = -5A^2
which doesn't make sense to me intuitively since the same variable is on both sides. I'm a bit confused, any help is appreciated.
A = -1/10 * dA/dt
dA/A=-10dt
lnA=-10t+C
when t=0, A=5
ln5=C
LnA=-10t+ln5
ah, I see! thank you
To solve the given differential equation, we can use separation of variables.
Starting with the differential equation:
dA/dt = -10A
Let's rearrange the equation by dividing both sides by A:
1/A dA = -10 dt
Now, integrate both sides of the equation with respect to their respective variables:
∫ (1/A) dA = ∫ (-10) dt
ln|A| = -10t + C
Where C is the integration constant.
To find the value C, we can use the initial condition A(0) = 5. Plugging this into the equation:
ln|5| = -10(0) + C
ln|5| = C
Therefore, we have:
ln|A| = -10t + ln|5|
Exponentiating both sides to eliminate the logarithm:
|A| = e^(-10t + ln|5|)
Since A is a positive quantity, we can drop the absolute value signs:
A = e^(-10t + ln|5|) = 5e^(-10t)
So, the solution to the given differential equation with the initial condition A(0) = 5 is:
A(t) = 5e^(-10t)
To solve the differential equation dA/dt = -10A with the initial condition A(0) = 5, you can use separation of variables.
First, let's rewrite the equation:
dA/A = -10 dt
Now, we can integrate both sides:
∫ (1/A) dA = ∫ (-10) dt
The integral of (1/A) with respect to A is ln|A| + C1, where C1 is the constant of integration. And the integral of -10 with respect to t is -10t + C2, where C2 is another constant of integration.
So, we have:
ln|A| + C1 = -10t + C2
Now, we can simplify by combining the two constants into a single constant:
ln|A| = -10t + C
Where C = C2 - C1 is the new constant.
Next, we can remove the natural logarithm by exponentiating both sides:
e^(ln|A|) = e^(-10t + C)
Simplifying further:
|A| = e^(-10t) * e^C
Since e^C is just another constant, let's combine it into a new constant D:
|A| = De^(-10t)
Now, we can handle the absolute value by splitting the equation into two cases:
Case 1: A > 0
If A is positive, then |A| = A. So we have:
A = De^(-10t)
Case 2: A < 0
If A is negative, then |A| = -A. So we have:
-A = De^(-10t)
Now, let's consider the initial condition A(0) = 5:
Case 1: A > 0
Substituting A = 5 and t = 0 into A = De^(-10t), we get:
5 = De^0
5 = D
So the solution for this case is:
A = 5e^(-10t)
Case 2: A < 0
Substituting A = -5 and t = 0 into -A = De^(-10t), we get:
-(-5) = De^0
5 = D
So the solution for this case is:
A = -5e^(-10t)
Thus, the general solution for the given initial condition A(0) = 5 is:
A = 5e^(-10t) for A > 0
A = -5e^(-10t) for A < 0