Find the center of mass of a lamina in the shape of an isosceles right triangle with equal sides of length $ a $ if the density at any point is proportional to the square of the distance from the vertex opposite the hypotenuse.
$$\begin{array}{l}{y^{\prime \prime}(t)=\cos (t-y)+y^{2}(t)} \\ {y(0)=1, \quad y^{\prime}(0)=0}\end{array}$$
$$\begin{array}{l}{y^{(6)}(t)=\left[y^{\prime}(t)\right]^{3}-\sin (y(t))+e^{2 t}} \\ {y(0)=y^{\prime}(0)=\cdots=y^{(5)}(0)=0}\end{array}$$
$$\begin{array}{ll}{3 x^{n}+5 x-2 y=0 ;} & {x(0)=-1, \quad x^{\prime}(0)=0} \\ {4 y^{\prime \prime}+2 y-6 x=0 ;} & {y(0)=1, \quad y^{\prime}(0)=2}\end{array}$$
Sturm Liouville Form. A second-order equation is said to be in Sturm Liouville form if it is expressed as$$\left[p(t) y^{\prime}(t)\right]^{\prime}+q(t) y(t)=0$$
SturmLiouville Form. A second-order equation is said to be in SturmLiouville form if it is expressed as$$\left[p(t) y^{\prime}(t)\right]^{\prime}+q(t) y(t)=0$$Show that the substitutions $x_{1}=y, x_{2}=p y^{\prime}$ result inthe normal form$$\begin{aligned} x_{1}^{\prime} &=x_{2} / p \\ x_{2}^{\prime} &=-q x_{1} \end{aligned}$$
