Another differential equation question.?

Question

I have a proof problem with the differential equation being y'=x+4sinx.....the solution is y=(x^2/2)-4cosx+2. Any advice? Thanks a million!

Why does "C" have to equal 2?

Answer 1

y=(x^2/2)-4cosx+2
y'=2x/2-4(-sinx)+0

y'=x+4sinx


if y'=x+4sinx
take integral
y=integral x dx+4 integral sinx dx
y=x^2/2+4(-cosx)+C

Let C=2 then
y=(x^2/2)-4cosx+2

Answer 2

To prove that a given f(x) is a solution of a diff. eqn, simply insert it into the given diff. eqn.

e.g. Suppose you have ...

af''(x) + bf'(x) + g(x) = c ( which is an example of a 2nd order eqn., effectively the same as yours but with a=0, b=1,c=0 )

Then simply differentiate your given soln. ( f(x) ) and verify that it satisfies your given diff. eqn.
No integration is necessary :-)

.....
P.S. Your C doesn't have to = 2, this is just a particular solution, the general soln. would just simply say "d" for example

Answer 3

first carry close the biggest factors complete salt answer is 200 litres intial concentration is 20g/l new answer further is 10l/hour which has concentration 5g answer bumped off is 10l/hour so intial concentration of salt interior the salt answer is 20g/litre in 2 hundred litres that's 20 * 2 hundred g = 4000g as quickly as a clean answer is further say after a time 't' the concentration of salt would be, further at 5g/litre at 10 litre in keeping with hour so 50g in keeping with hour i.e 50t (say after a time t) bumped off is 10 litres, so at any time t the somewhat concentration of salt would be (4000 + 50t) in 210 litres so if 10 litre is bumped off the quantity of salt bumped off would be (4000 + 50t)/210 * 10 so the desirable equation of S(t) would be 4000g intial salt 50t in keeping with further in keeping with t (4000 + 50t)/21 bumped off in keeping with t 4000 + 50t - (4000 +50t)/21 S(t) = (4000 + 50t) (20/21)