# If f(x+y) + f(x-y) =2f(x)f(y) and f(0)=k then find f(x)?

• succhi
In summary, the given equation is f(x+y) + f(x-y) = 2f(x)f(y) and the initial condition is f(0) = k. The general solution for f(x) is f(x) = c*cos(x) + d*sin(x), where c and d are constants. To find the specific solution for f(x) with the given initial condition, we can substitute c = k and d = f'(0) into the general solution. This equation and initial condition can have infinitely many solutions due to the infinite possible values for c and d.

#### succhi

if
f(x+y) + f(x-y) =2f(x)f(y)
and f(0)=k
then find f(x)?

What should be approach followed to solve this question .

You have an expression with f(x) and f(y) and you only want f(x). So you have to get rid of y.

One more hint: that identity holds for all values of y.

## 1. What is the given equation and initial condition?

The given equation is f(x+y) + f(x-y) = 2f(x)f(y) and the initial condition is f(0) = k.

## 2. What is the general solution for f(x)?

The general solution for f(x) is f(x) = c*cos(x) + d*sin(x), where c and d are constants.

## 3. How do you use the initial condition to find the values of c and d?

Substituting x = 0 and f(0) = k into the general solution, we get k = c*cos(0) + d*sin(0) = c. This means that c = k. Substituting this into the general solution, we get f(x) = k*cos(x) + d*sin(x). To find d, we can differentiate both sides of the equation with respect to x and substitute x = 0 to get d = f'(0).

## 4. How do you find the specific solution for f(x) given the initial condition?

Substituting c = k and d = f'(0) into the general solution, we get f(x) = k*cos(x) + f'(0)*sin(x). This is the specific solution for f(x) with the given initial condition.

## 5. Can this equation and initial condition have multiple solutions?

Yes, this equation and initial condition can have infinitely many solutions because there are infinitely many possible values for c and d. Each pair of values for c and d will result in a different solution for f(x).