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 $z.$
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 $$u(z)=\sqrt{\frac{\pi}{2}}\cdot \text{erfi}\left(\frac{z}{\sqrt{2}}\right)\cdot e^{-\frac{z^2}{2}}+e^{-\frac{z^2}{2}}$$