MCQ
Solution of differential equation $\frac{{dy}}{{dx}} = 2xy$ is
- ✓$y = c{e^{{x^2}}}$
- B${y^2} = 2{x^2} + c$
- C$y = {e^{ - {x^2}}} + c$
- D$y = {x^2} + c$
==> $y = {e^{{x^2} + c}} = {e^c}{e^{{x^2}}}$ ==> $y = c{e^{{x^2}}}$.
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$(1,1)(-1,1),(1,-1),(-1,-1),(-2,1)(2,-1),(-1,2)$ and $(-2,-1)$