I found general solution of this eq: $$\frac {dy}{dx}=5xy$$
General solution: $$y(x)=e^{5x^2/2+c}$$
Now, i want to check my answer in first eq: $$5xe^{5x^2/2}=5xe^{5x^2/2+c}$$
If $c\neq0$ for some values like y(1)=1, c=-5/2, then equation is incorrect.
Where do I think wrong?
Note that $$\frac {dy}{dx}=\frac {d }{dx}\,e^{5x^2/2+c}=5xe^{5x^2/2+c}$$ $$\frac {dy}{dx}=5xy(x)$$