We are asked to solve this using Variation of Parameters (VoP), given: \tag 1 y''-3y'+2y=4x+4 Step 1 Find the homogenous solution to (1), so we have: \tag 2 y''-3 y'+ 2 y = 0 This yields: y_h = c_1e^x + c_2 e^{2x} ...

Assuming a dense system, solving the KKT system gives the solution as \begin{bmatrix} Q & A^*\\A & 0 \end{bmatrix}\begin{bmatrix} \mathbf{x}^\star\\\boldsymbol{\nu}^\star \end{bmatrix} = -\begin{bmatrix} \mathbf{p}\\\mathbf{b} \end{bmatrix} ...

Hint: You are missing an equilibrium point. \dot{x}=0=x^2-y^2 \implies x=\pm y \dot{y}=0=x(1-y) \implies x=0 \text{ or } y=1 Using x=0 for the first equation we see that y=0, which implies ...

Use the Divergence Theorem for the whole thing. This converts the surface integral to a volume integral: \iint_T F\cdot \hat{n} dS = \iiint_V (\nabla \cdot F) dV This is much more doable to ...

I think it must be 1+m-p=1 and mp=h^4 solving this we get m=\pm h^2 and p=\pm h^2

Since you asked for an intuitive way to understand covariance and contravariance, I think this will do. First of all, remember that the reason of having covariant or contravariant tensors is because ...