a) Compute the Wronskian and simplify
\begin{align} y_1 = t^2 + 1\\ y_2 = 3t^2 + k \end{align}
So I computed the Wronskian and got: \begin{align}6t-2kt \Longleftrightarrow t(6-2k). \end{align}
b) For what values of $k$ are the two functions linearly independent?
So I know that the functions are linearly independent because I took the Wronskian and didn't get $0$. I'm not sure though for what values of $k$.. Would it be all values of $k$ as long as $k$ is not $3$ because $6t-6t = 0$ ?
The Wronskian is \begin{align} W\left[y_1,\cdots y_n\right]\left(x\right):=\det\begin{bmatrix} y_1 & y_2 & \cdots & y_n\\ y'_1 & y'_2 & \cdots & y'_n\\ \vdots & \vdots & \ddots & \vdots\\ y^{\left(n-1\right)}_1 & y^{\left(n-1\right)}_2 & \cdots & y^{\left(n-1\right)}_n \end{bmatrix}, \end{align} and using your functions we have \begin{align} y_1=t^2+1,\\y_2=3t^2+k, \end{align} which gives us \begin{align} W\left[y_1,y_2\right]\left(t\right)&=\det\begin{bmatrix} t^2+1 & 3t^2+k\\ 2t & 6t \end{bmatrix}\\ &=6t\left(t^2+1\right)-\left(3t^2+k\right)2t\\ &=6t^3+6t-\left(6t^3+2tk\right)\\ &=6t-2tk, \end{align} and therefore the linear independence of these two functions is dependent upon $k\neq 3$, otherwise these two solutions would not be linearly independent.