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

Answers : (2)

Sujit Kumar
111 Points
							
 
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.
one year ago
Aditya Gupta
1714 Points
							
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
one year ago
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