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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

```							 f(x)=a0+a1x+a2x2+a3x3+.....anxnf(0)=a0        also given.... f(0)=2      Therefore a0=2f’(x)=a1+2a2x+3a3x2+......nanxn-1Therefore 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 a5You can go on until you get a pattern.
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one year 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)= yWe are given thaty"=yOr dy'/dx=yOr (dy'/dy)*(dy/dx)=yBut dy/dx= y'So 2y'dy'= 2ydyIntegrating both sidesy'^2= y^2+CPut x=0So 9= 4+CC=5So y'= √(y^2+5)Or dy/√(y^2+5)= dxx+K= 1/√5 arctan y/√5Now again put x=0 to obtain the value of K as 1/√5 arctan 2/√5
```
one year ago
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