I am trying to solve the following problem
\begin{align*}
u_{t}-\frac{v}{2}\cdot u_{xx}+\frac{v}{2}\cdot x^2 \cdot u(x)&=0 \\
u(x,0)&=f(x)
\end{align*}
Where $f(x)=y_1(x)+0.2y_4(x)+0.01y_6(x)$.
$y_n$ are the Eigenfunctions that defined as $y_n(x)=\exp\left(\displaystyle{\frac{-x^2}{2}}\right)H_n(x)$
By separation of we assume $$u(x,t)=X(x)T(x)$$
I have found that the general solution of $T$ is $$T(x)=c_1e^{-\lambda v t}$$ and the general solution of $X$ is $$X_n(x)=\exp\left(\displaystyle{\frac{-x^2}{2}}\right)H_n(x)$$
So the general solution to our initial problem is $$u_n(x,t)=\sum_{n=1}^{\infty}A_n \exp\left(\displaystyle{\frac{-x^2}{2}}\right)H_n(x) c_1e^{-\lambda v t}$$
Using the I.C we got $$u(x,0)=\sum_{n=1}^{\infty}A_n \exp\left(\displaystyle{\frac{-x^2}{2}}\right)H_n(x)=f(x)$$ How can I write the result in Fourier series form?
To finish your solution, you need to solve the equation $$ \sum_{n=1}^\infty A_n y_n(x) = y_1(x) + \frac{1}{5}y_4(x) + \frac{1}{100} y_6(x) $$ for the $A_n$. To do this, you need to use the orthogonality property of hermite polynomials: $$ \int_{-\infty}^\infty H_m(x) H_n(x) e^{-x^2/2} dx = \sqrt{2\pi}n! \delta_{nm}.$$ Note we can rewrite this as $$ \int_{-\infty}^\infty H_m(x) y_n(x) dx = \sqrt{2\pi}n! \delta_{nm}.$$ So we multiply the first equation by $H_m(x)$ and integrate over $x$. $$ \sum_{n=1}^\infty A_n \sqrt{2\pi}n! \delta_{nm} = \sqrt{2\pi}1! \delta_{1m} + \frac{1}{5}\sqrt{2\pi}4! \delta_{4m}+ \frac{1}{100}\sqrt{2\pi}6! \delta_{6m} .$$ Can you take it from here? You just need to get $A_n$ for each $n=1,2,\dots,$