Question

Find the solution to the differential equation dx/dt=2x+2t+tx+4 that satisfies the conditions x>=0 and x(1)=2

Answer 1

i cant even tell if im reading it right

Answer 2

combine all your "x" terms

Answer 3

they will tell you the function associated with "x"

Answer 4

Notice that
2x+2t+tx+4 = (x+2)(t+2)

Answer 5

x' =(2+t)x+(2t+4)

x'-(2+t)x =2t+4

might be a way to format it to a usual look

Answer 6

dunno if its best, but at least it gets you to a "pattern" that normal

Answer 7

Look ... because
dx/dt = 2x+2t+tx+4 = (x+2)(t+2)

this equation is separable:

\[ \frac{dx}{x+2} = (t+2) dt \]
Now integrate

Answer 8

separable is soooo last chapter tho :)

Answer 9

that was weird

Answer 10

@Lammy, making sense?

Answer 11

let me try

Answer 12

Ln(x+2)+C = 1/2t^2+2t+c

Answer 13

C=2ln2-(5/2)
so ln(x+2)=1/2t^2+2t+2ln(2)-(5/2)

Answer 14

only one side needs a +c ; as a shortcut; then solve for x

Answer 15

ln(x+2) = 1/2 t^2 + 2t + c

x+2 = exp(1/2 t^2 + 2t + c)

x = exp(1/2 t^2 + 2t + c) - 2

x = Cexp(1/2 t^2 + 2t) - 2

Answer 16

i dont have to solve for C first?

Answer 17

You can solve for C now using the initial condition provided to you in the problem:
x(1) = 2.

So take the general solution

\[ x(t) = Ce^{\frac{t^2}{2} + 2t} - 2 \]
and now find C.