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GROUPS:
I am suppose to solve (fg)'=f'g' given f(x)=e^(x^2). I am suppose to find g(x)
POSTED BY: Joe Morrow
Answer
6 months ago
In[1]:= DSolve[D[Exp[x^2]*g[x], x] == Exp[x^2] g'[x], g[x], x]

Out[1]= {{g[x] -> 0}}
POSTED BY: Frank Kampas
Answer
6 months ago
f[x_] := Exp[x^2]

DSolve[D[f[x] g[x], x] == f'[x] g'[x], g[x], x] // Simplify

{{g[x] -> E^(-(1/2) + x) Sqrt[-1 + 2 x] C[1]}}
Since E^(-(1/2)+x) can be rewritten as e^-(1/2) e^x , the e^-(1/2) can be included within the arbitrary constant and the solution is 

g =  C_1 e^x Sqrt[2x-1]
POSTED BY: Melvyn Stancliffe
Answer
6 months ago
 In[3]:= f[x_] := Exp[x^2]
 
 
 In[4]:= DSolve[f[x] g'[x] + f'[x] g[x] == f'[x] g'[x], g[x], x]
 
 
 Out[4]= {{g[x] -> E^(1/2 (-1 + 2 x + Log[-1 + 2 x])) C[1]}}
 
 
In[5]:= Simplify[%]


Out[5]= {{g[x] -> E^(-(1/2) + x) Sqrt[-1 + 2 x] C[1]}}
POSTED BY: S M Blinder
Answer
6 months ago