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Find a particular solution that satasfies the D.E f "(x)= 2x and its initial conditions f '(1)= 2, f(0)= -1

zzzzz  Aug 4, 2017
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f " (x)  = 2x   f ' (1)   = 2, f (0)  = -1

 

f "(x)   = 2x        integrate to find the first derivative

 

f ' (x) =  x^2  + C     applying the initial condition to find C1 ,  we have that

 

f '( 1)  = 2

 

2  =    (1)^2 + C1

 

2  = 1 + C1

 

1  = C1

 

So....  f'(x) =  x^2 + 1

 

Integrate this again to find y(x)

 

f(x)  = (1/3)x^2 + x  + C2

 

And applying  the  second initial condition to find C2, we have that

 

f(0) = -1

 

-1 = (1/3)(0)^3 + 0 + C2

 

C2  = -1

 

So.....   f(x)  =  (1/3)x^3 +  x   - 1

 

 

 

cool cool cool

CPhill  Aug 4, 2017

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