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The general solution of $\large\frac{dy}{dx}$$=2xe^{x^2-y}$ is: \[(A)\;e^{x^2-y}=c \quad (B)\;e^{-y}+e^{x^2}=c \quad(C)\;e^y=e^{x^2}+c \quad (D)\;e^{x^2}+y=c\]

1 Answer

  • Linear equation of the type $\large\frac{dy}{dx}$$=f(x)$ can be solved by seperating the variable, and then integrating
  • $\int e^x dx=e^x+c$
Given $\large\frac{dy}{dx}$$=2xe^{x^2-y}$
This can be written as :$\large\frac{dy}{dx}=\frac{\Large2xe^{x^2}}{e^y}$
Now seperating the variables we get,
Now integrating on both sides,
$\int e^y.dy=\int 2xe^{x^2}.dx$
Put $x^2=t$ on differentiating w.r.t x we get $2xdx=dt$
Now substituting this we get
$\int e^y.dy=\int e^t.dt$
Substitute for $t$ we get
Hence the correct option is $C$
answered May 19, 2013 by meena.p

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