if f'(x)=2f(x) and f(2)=1, then f(x)= ?
f'(x)=2f(x)
f'(x)/f(x) = 2
substitute u=f(x) and integrate both sides,
du/u = 2
ln(u)=2x + C
u=e^(2x+C)
f(x)=e^(2x+c)
f(2)=e^(2*2+C)=1 => C=-4
So f(x)=e^(2x-4)
Well, if f'(x) = 2f(x), that means the rate of change of f(x) is twice the value of f(x) itself. It's like f(x) has a case of multiplication madness!
Now, let's solve the problem using humor: Imagine f(x) is a plant trying to grow exponentially. It starts at f(2) = 1, but due to this multiplication madness, its growth rate doubles every step of the way.
So, let's see how this plant-gone-crazy develops over time:
f(2) = 1
f'(2) = 2 * f(2) = 2 * 1 = 2
At x = 3, it grows as twice as fast: f'(3) = 2 * f(3) = 2 * f'(2) = 2 * 2 = 4
At x = 4, it grows even faster: f'(4) = 2 * f(4) = 2 * f'(3) = 2 * 4 = 8
As you can see, the growth rate keeps doubling and becomes quite exponential. Therefore, f(x) = 2^x.
So, clownish conclusion, f(x) = 2^x! Your plant has turned into a wild exponential adventure!
To find the function f(x), we can solve the given differential equation f'(x) = 2f(x). This is a first-order linear ordinary differential equation.
First, let's separate the variables by dividing both sides by f(x):
f'(x)/f(x) = 2
Next, let's integrate both sides with respect to x:
∫f'(x)/f(x) dx = ∫2 dx
Using the property of logarithms, the integral of 1/u du is ln|u|. Applying this property to the left side of the equation gives:
ln|f(x)| = 2x + C1,
where C1 is the constant of integration.
Now, let's exponentiate both sides:
e^(ln|f(x)|) = e^(2x + C1)
Since e^(ln|f(x)|) cancels each other out, we get:
|f(x)| = e^(2x + C1)
Now we can remove the absolute value by splitting into two cases:
Case 1: f(x) > 0:
In this case, we have f(x) = e^(2x + C1).
Case 2: f(x) < 0:
In this case, we have f(x) = -e^(2x + C1).
To determine the constant of integration C1, we use the initial condition f(2) = 1. Substituting x = 2 and f(x) = 1 into our solutions, we can solve for C1:
Case 1:
f(2) = e^(2*2 + C1) = e^(4 + C1) = 1.
Taking the natural logarithm of both sides:
ln(e^(4 + C1)) = ln(1)
4 + C1 = 0
C1 = -4
Case 2:
f(2) = -e^(2*2 + C1) = -e^(4 + C1) = 1.
Taking the natural logarithm of both sides:
ln(-e^(4 + C1)) = ln(1)
This equation implies that there is no solution for C1 when f(x) < 0.
Therefore, the function f(x) when f(2) = 1 is:
f(x) = e^(2x - 4).
To find the function f(x) given that f'(x) = 2f(x) and f(2) = 1, we can use the technique of solving a first-order linear ordinary differential equation.
Step 1: Separate the variables.
f'(x) = 2f(x)
Divide both sides by f(x):
f'(x) / f(x) = 2
Step 2: Integrate both sides with respect to x.
∫ f'(x) / f(x) dx = ∫ 2 dx
The left side integrates to:
∫ f'(x) / f(x) dx = ln|f(x)| + C1, where C1 is the constant of integration.
The right side integrates to:
∫ 2 dx = 2x + C2, where C2 is another constant of integration.
Step 3: Solve for f(x).
Putting everything together, we have:
ln|f(x)| + C1 = 2x + C2
Using the properties of logarithms, we can rewrite the equation as:
ln|f(x)| = 2x + (C2 - C1)
Now, applying the exponentiation function (e^x) to both sides, we get:
|f(x)| = e^(2x + (C2 - C1))
Since e^(C2 - C1) is just another constant, let's represent it as C3:
|f(x)| = C3 * e^(2x)
Given the initial condition f(2) = 1, we substitute x = 2 into the equation:
|f(2)| = C3 * e^(2*2)
|1| = C3 * e^4
1 = C3 * e^4
Since |f(x)| represents the absolute value of f(x), we have two possibilities:
1) Let's consider f(x) = C3 * e^(2x), where C3 = e^4:
f(x) = e^4 * e^(2x) = e^(4 + 2x)
2) Alternatively, let's consider f(x) = -C3 * e^(2x), where C3 = e^(-4):
f(x) = -e^(-4) * e^(2x) = -e^(-4 + 2x)
So, the two possible solutions for f(x) are:
1) f(x) = e^(4 + 2x)
2) f(x) = -e^(-4 + 2x)