Calc differential equations

I need a step by step explabation on how to' do this so I can undersrand please
1)
Verify that 'y' is a general solution to' the differential equation, Find a particular solution of the diff equation w/side condition. Y'-2 (y/x) =2;
Y=cx^2-2x;
Y(1)=10

2) A bucket initially has no sand, and A(t) represents amnt of sand present in The bucket at any time t . Sand flows into The bucket and out of The bucket by The differential equation A/t =30- A/5. How many sand is present at The end of 10 minutes?
How much sand is present in The long run?

3) The population (square miles) of a town density is f(x)=5000xe^y / 1+2x^2 (0< x< 4; -2 x and y Are in miles
What is pop inside area by R=. {x,y}| 0< x < 4); -2< y < 0)}?
What is average pop in The area(per sq. Mile) (2)(4) = 8 sq. Miles.

1.) y' - 2y/x = 2

Integrating factor = e^(∫-2/x dx) = 1/x^2

y'/(x^2) - 2y/(x^3) = 2/x^2

d/dx[y/x^2] = 2/x^2

y/x^2 = -2/x + C

y = -2x + Cx^2

Plugging in initial condition:

10 = -2 + C => C = 12

y = -2x + 12x^2

2.) Did you mean.. dA/dt = 30 - A/5 ?

If so, then:

5 dA/(150 - A) = dt

Integrating both sides:

-5*ln|150 - A| = t + C

150 - A = C*e^(-t/5)

A = 150 - C*e^(-t/5)

Since the bucket initially has no sand, so A(0) = 0.

0 = 150 - C => C = 150

A = 150(1 - e^(-t/5))

At the end of 10 minutes, the amount of sand present is A(10) = 150(1 - e^(-2))

In the long run, the amount of sand present is lim t->inf. A(t) = 150

Note that units for time were mentioned here and no units for the volume of sand was mentioned.

3.) f(x,y) = 5000x*e^(y)/(1 + 2x^2)

Total population inside area = ∫[-2, 0] ∫[0, 4] 5000x*e^(y)/(1 + 2x^2) dx dy

= ∫[-2, 0] 1250*e^(y)*ln(1 + 2x^2) eval. from x = 0 to x = 4 dy

= 1250*∫e^(y)*(ln(33)) dy from -2 to 0

= 1250*ln(33)*(1 - e^(-2))

People per square mile = 156.25 ln(33)*(1 - e^(-2))