Solve the initial value problem dzdx=2e2x\frac{dz}{dx} = 2e^{2x}dxdz=2e2x with z(0)=4z(0) = 4z(0)=4.
z=e2x+3z = e^{2x} + 3z=e2x+3
z=2e2x+2z = 2e^{2x} + 2z=2e2x+2
z=e2x+4z = e^{2x} + 4z=e2x+4
z=e2x+5z = e^{2x} + 5z=e2x+5