Use the Laplace transform to solve the following

Find the orthogonal trajectory.
${\displaystyle y^2=k{x}^3}$

Find a general solution ${\displaystyle y\left(x\right)}$ to the differential equation ${\displaystyle y^{\prime }=\frac{y}{x}+2x^2}$

Let $y(t)={\int }_0^{t}f(t)dt$ If the Laplace transform of y(t) is given $Y(s)=\frac{19}{(s^2+25)}$ , find f(t)
a) $f(t)=19\sin (5t)$
b) none
c) $f(t)=6\sin (2t)$
d) $f(t)=20\cos (6t)$
e) $f(t)=19\cos (5t)$

A question asks us to solve the differential equation
${\displaystyle -u\left(x\right)=\delta \left(x\right)}$
with boundary conditions
${\displaystyle u(-2)=0\text{and}u\left(3\right)=0}$ where ${\displaystyle \delta \left(x\right)}$ is the Dirac delta function. But inside the same question, teacher gives the solution in two pieces as ${\displaystyle u=A(x+2)\text{for}x\le 0\text{and}u=B(x-3)\text{for}x\ge 0}$. I understand when we integrate the delta function twice the result is the ramp function R(x). However elsewhere in his lecture the teacher had given the general solution of that DE as
${\displaystyle u\left(x\right)=-R\left(x\right)+C+Dx}$

Find the inverse laplace transform of the function
$Y(s)=\frac{e^{-s}}{s(2s-1)}$

use the inverse form of the First translation theorem to find the following laplace transform
$L^{-1}\left\{\frac{s}{s^2+4s+5}\right\}$

Solve the question with laplace transformation
$y^{″}+2y^{\prime }+y=4$
$y(0)=3$
$y^{\prime }(0)=0$