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If f(0)=2, f'(0)=3, f"(x)=f(x) then what is the value of f(x)?

 
If f(0)=2, f'(0)=3, f"(x)=f(x) then what is the value of f(x)?

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2 Answers

Sujit Kumar
111 Points
5 years ago
 
f(x)=a0+a1x+a2x2+a3x3+.....anxn
f(0)=a0        also given.... f(0)=2
      Therefore a0=2
f’(x)=a1+2a2x+3a3x2+......nanxn-1
Therefore f’(0)=a1
     also given.... f’(0)=3
    Therefore a1=3
 
Now you can double differentiate f(x)=f’’(x) and get the values of a2=1 and a3=0.5 and input them in the original equation f(x) and now a2 and a3 have to be equal to a4 and a5
You can go on until you get a pattern.
Aditya Gupta
2081 Points
5 years ago
The above answer is an absolute mess and nothing more than an absurdity. So lemme tell you the correct method of solving it.
Let f(x)= y
We are given that
y"=y
Or dy'/dx=y
Or (dy'/dy)*(dy/dx)=y
But dy/dx= y'
So 2y'dy'= 2ydy
Integrating both sides
y'^2= y^2+C
Put x=0
So 9= 4+C
C=5
So y'= √(y^2+5)
Or dy/√(y^2+5)= dx
x+K= 1/√5 arctan y/√5
Now again put x=0 to obtain the value of K as 1/√5 arctan 2/√5

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