Power series solutions 
Construct two linearly independent, power series solutions to the ODE
  $$u''+zu'+u=0.$$
  Hence find the solution which satisfies $u(0)=1$ and $u'(0)=1.$

I have come up with the solution for the coeffecient, however I am not sure why we multiply them with $z^2$ instead of $z.$ 
 A: Let us consider your differential equation:
$$\frac{d^2u(z)}{dz^2}+z\cdot\frac{du(z)}{dz}+u(z)=0$$
Apply the reverse product rule:
$$\frac{d}{dz}\frac{du(z)}{dz}+\frac{d}{dz}(z\cdot u(z))=0$$
Integrate with respect to $z$:
$$\int\frac{d}{dz}\left(\frac{du(z)}{dz}+z\cdot u(z)\right) dz=0$$
$$\frac{du(z)}{dz}+z\cdot u(z)=C_1$$
Multiply on both sides by $e^{\frac{z^2}{2}}$:
$$e^{\frac{z^2}{2}}\cdot\frac{du(z)}{dz}+z\cdot e^{\frac{z^2}{2}}\cdot u(z)=C_1\cdot e^{\frac{z^2}{2}}$$
Recognize that $z\cdot e^{\frac{z^2}{2}}=\frac{d}{dz}\left(e^{\frac{z^2}{2}}\right)$ and apply the reverse product rule:
$$\frac{d}{dz}\left(e^{\frac{z^2}{2}}\cdot u(z)\right)=C_1\cdot e^{\frac{z^2}{2}}$$
Integrate with respect to $z$:
$$\int\frac{d}{dz}\left(e^{\frac{z^2}{2}}\cdot u(z)\right) dz=\int C_1\cdot e^{\frac{z^2}{2}} dz$$
$$e^{\frac{z^2}{2}}\cdot u(z)=C_1\cdot\sqrt{\frac{\pi}
{2}}\cdot \text{erfi}\left(\frac{z}{\sqrt{2}}\right)+C_2$$
Multiply on both sides by $e^{-\frac{z^2}{2}}$:
$$u(z)=C_1\cdot\sqrt{\frac{\pi}{2}}\cdot \text{erfi}\left(\frac{z}{\sqrt{2}}\right)\cdot e^{-\frac{z^2}{2}}+C_2\cdot e^{-\frac{z^2}{2}}$$
The first derivative is given by
$$\frac{du(z)}{dz}=C_1-C_1\cdot \sqrt{\frac{\pi}{2}}\cdot \text{erfi}\left(\frac{z}{\sqrt{2}}\right)\cdot z \cdot e^{-\frac{z^2}{2}}-C_2\cdot z \cdot e^{-\frac{z^2}{2}} $$
Apply conditions:
$$u(0)=1\rightarrow u(0)=C_2\rightarrow C_2 =1$$
$$u'(0)=1\rightarrow u'(0)=C_1\rightarrow C_1=1$$
Thus, a solution is given by
$$u(z)=\sqrt{\frac{\pi}{2}}\cdot \text{erfi}\left(\frac{z}{\sqrt{2}}\right)\cdot e^{-\frac{z^2}{2}}+e^{-\frac{z^2}{2}}$$
