Let \( \mathrm{y}^{\prime}= \) Ay be a system of differential equations where \( A=\left[\begin{array}{cc}-5 & 1 \\ 0 & -5\end{array}\right] \), The matrix has spectrum \( \lambda(A)=\{-5\} \) and \( \mathbf{c}=\left[\begin{array}{l}5 \\ 0\end{array}\right] \) is an eigenvector of \( A \) corresponding to \( r=-5 \), The vector \( \mathrm{d}=\left[\begin{array}{c}-3 \\ 5\end{array}\right] \) satisfies the equation \( (A-r I) \mathrm{d}=\mathbf{c} \). What is the general solution to the system of differential equations? \( \left[\begin{array}{l}y_{1} \\ y_{2}\end{array}\right]=k_{1} e \)

Find the volume of the solid obtained by rotating the region bounded by the curves \( y = x^{2} \) and \( y = 4 \) about the line \( y = 5 \) using the washer method.

(d) \( \lim _{x \mapsto-\infty} \frac{x^{5}+x^{4}}{x^{3}-x} \)

a- Find the second derivative for the following function: \[ x(t)=\sin ^{-1}(t), \quad y(t)=\sqrt{1-t^{2}} \]

(f) \( \lim _{x \mapsto 0^{-}} \frac{-2 \cdot x^{2}}{x^{4}+x^{3}} \)

Paraffin is stored in a tank with a horizontal base. At time \( t \) minutes, the depth of paraffin in the tank is \( x \mathrm{~cm} \). When \( t=0, x=72 \). There is a tap in the side of the tank through which the paraffin can flow. When the tap is opened, the flow of the paraffin is modelled by the differential equation \[ \frac{\mathrm{d} x}{\mathrm{~d} t}=-4(x-8)^{\frac{1}{3}} \text {. } \] (i) How long does it take for the level of paraffin to fall from a depth of 72 cm to a depth of 35 cm ? (ii) The tank is filled again to its original depth of 72 cm of paraffin and the tap is then opened. The paraffin flows out until it stops. How long does this take?