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$If f^(')(x)=2f(x) and f(1)=5, then f(3)=me^(n) for some integers m and n. What are m and n ? {:[m=],[n=]:}$

If $f^{\prime}(x)=2 f(x)$ and $f(1)=5$ , then $f(3)=m e^{n}$ for some integers $m$ and $n$ . $\newline$ What are $m$ and $n$ ? $\newline$ $\begin{array}{l} m= \square \\ n= \square \end{array}$

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Math Problems

Calculus

Find derivatives of sine and cosine functions

Full solution

Q. If

f^{\prime}(x)=2 f(x)

and

f(1)=5

, then

f(3)=m e^{n}

for some integers

m

and

n

\newline

What are

m

and

n

\newline

\begin{array}{l} m= \square \\ n= \square \end{array}

Recognize Differential Equation: First, we recognize that $f'(x) = 2f(x)$ is a differential equation that describes exponential growth or decay. Since $f(1) = 5$ , we can use this initial condition to find the particular solution.
Guess Exponential Form: We can guess that $f(x)$ is of the form $Ce^{kx}$ because the derivative of an exponential function is proportional to the function itself. So, $f'(x) = kCe^{kx}$ .
Substitute and Solve: Substituting $f(x) = Ce^{kx}$ into the differential equation $f'(x) = 2f(x)$ , we get $kCe^{kx} = 2Ce^{kx}$ . Dividing both sides by $Ce^{kx}$ , we find that $k = 2$ .
Find Constant $C$ : Now we know that $f(x) = Ce^{2x}$ . We can use the initial condition $f(1) = 5$ to find $C$ . Substituting $x = 1$ , we get $f(1) = Ce^{2\cdot 1} = 5$ .
Calculate Particular Solution: Solving for $C$ , we have $Ce^{2} = 5$ . Dividing both sides by $e^{2}$ , we find $C = \frac{5}{e^{2}}$ .
Find $f(3)$ : So the particular solution to the differential equation is $f(x) = \left(\frac{5}{e^{2}}\right)e^{2x}$ . Now we need to find $f(3)$ .
Find $f(3)$ : So the particular solution to the differential equation is $f(x) = \frac{5}{e^{2}}e^{2x}$ . Now we need to find $f(3)$ .Substituting $x = 3$ into $f(x)$ , we get $f(3) = \frac{5}{e^{2}}e^{2\cdot 3} = \frac{5}{e^{2}}e^{6}$ .
Find $f(3)$ : So the particular solution to the differential equation is $f(x) = \frac{5}{e^{2}}e^{2x}$ . Now we need to find $f(3)$ .Substituting $x = 3$ into $f(x)$ , we get $f(3) = \frac{5}{e^{2}}e^{2\cdot 3} = \frac{5}{e^{2}}e^{6}$ .Simplifying the expression, we have $f(3) = 5e^{4}$ . This is the same as $5e^{4n}$ where $n = 1$ , since $4\cdot 1 = 4$ .