Solve the initial value problem yy′+x=x2+y2√yy′+x=x2+y2 with y(3)=27−−√y(3)=27. to solve this, we should use the substitution u=u= (formulas) u′=u′= (formulas) enter derivatives using prime notation (e. g., you would enter y′y′ for dydxdydx). after the substitution from the previous part, we obtain the following linear differential equation in x, u,u′x, u,u′. (equations) the solution to the original initial value problem is described by the following equation in x, yx, y. (equations)

answer: are you

step-by-step explanation:

step-by-step explanation:

12: 00 p.m. to 7: 00 p.m. is a total of 7 hours.

4: 00 p.m. to 7: 00 p.m. is a total of 3 hours.

this means that from 4: 00 p.m to 7: 00 p.m. this represents 3/7 of what was lost from 12: 00 p.m. to 7: 00 p.m.

after knowing this, after doing 11,600 - 6,800 = 4,800

4,800 represents 3/7 of what was lost for 7 hours.

we then divide 4,800 by 3 which gets 1,600.

because we know that 1,600 represents 1/7 of how much water was lost, we just have to multiply by 7, which gets 11,200.   after adding 11,200 to how much water was left at 7: 00 p.m. which was 6,800. 11,200 + 6,800 = 18,000

this means that the pool originally had 18,000 gallons in the pool.

you can also check your work by doing 18,000 - 4(1600), which becomes           1800 - 6400 which gets you 11,600.