How do I solve y'= 4y, y(0)=1?

If  y'  =   4y   .....  then.......

dy / dx   =  4y       this is a separable equation

dy / y  =  4 dx      integrate both sides

∫ 1/y    dy     =   ∫ 4 dx

ln y    =   4x  + C          exponentiate both sides

e^(ln y)  =  e^(4x + C)  

y   =   e^C * e^(4x)  = Ce^(4x)

And we have that  y(0)  = 1 .....so.....

1  =  Ce^(4 * 0)

1  = C

And  when y  =4

  4 =  e^(4*x)       take the Ln  of both sides

Ln 4   =  Ln e^(4x)

Ln 4  = (4x) Ln e

Ln 4  =  4x     divide both sides by 4

Ln 4 / 4   =  x   ≈  .34657