Verify that $y=\frac{6}{5}-\frac{6}{5}{e}^{-20t}$ is the solution of the differential equation $\frac{dy}{dx}+20y=24.$

Solve the given differential equation $(2-x)dy-2ydx=0$

Determine whether the following are exact or not:

$($a$)(2x+y)dx-(x+6y)dy=0$

$($b$)(5y-2x)\frac{dy}{dx}-2y=0$

Solve this homogeneous differential equation $(x^2+y^2)dx+(x^2-xy)dy=0$

$5.$ Verify that the given functions form a fundamental set of solutions of the differential equation on the indicated interval. Form the general solution.

$($a$)y^{\text{'}\text{'}}-y^{\text{'}}-12y=0$;$e^{-3x},e^{4x},(-,).$

$($b$)x^2{y}^{\text{'}\text{'}}-6x{y}^{\text{'}}+12y=0$;$x^3,x^4,(0,).$

$6.$ Find the general solution of the given second$-$order differential equations:

$($a$)y^{\text{'}\text{'}}+8y^{\text{'}}+16y=0$

$($b$)y^{\text{'}\text{'}}-4y^{\text{'}}+5y=0$

$7.$ Solve the given differential equation by using undetermined coefficients:

$($a$)y^{\text{'}\text{'}}+3y^{\text{'}}-4y=2x^2-3x+1$

$($b$)y^{\text{'}\text{'}}-16y=2e^{4x}$

$8.$ Find the general solution the given differential equation by variation of parameters:

$($a$)y^{\text{'}\text{'}}-3y^{\text{'}}+2y=(x+3)e^{2x}$

$($b$)4y^{\text{'}\text{'}}-y=x{e}^{\frac{x}{2}}$ given the initial conditions $y(0)=1,y^{\text{'}}(0)=0.$

$9.$ Find the given inverse laplace transform

$($a$)L^{-1}\{\frac{(s+4)}{(s-3)(s+6)}\}$

$($b$)L^{-1}\{\frac{2s}{9s^2+1}\}$

$10.$ Find $F(s)$ or $f(t)$ as indicate in the following:

$($a$)L^{-1}\{\frac{5s}{(s-2{)}^2}\}$

$($b$)L\{t^{10}{e}^{-7t}\}$