Differential Equation: dQ/dx = -2Q where Q = 40 when x = 0

[calcidiot]

I need a little help with dQ/dx = -2Q where Q = 40 when x = 0

I don't think I am doing it correctly. Since I don't know the answer I have no idea if I am on the right track.

A)

dQ/dx = -2Q

y = dQ = -2Q dx and x = q

= -2[q^1/1] + 1[q^2/2] + C

= -2q + q^2 + C

B)

40 = -2(0) + (0)^2 + C

40 = C

Q = -2q + q^2 + 40


I understand that once you find the answer formula that you set it equal to 40 and plug 0 in for each x. But I don't think that I have part A correct.

 

[galactus]

Re: Differential Equation

Did you try separating variables?.

\(\displaystyle \frac{dQ}{dx}=-2Q\)

\(\displaystyle \frac{dQ}{Q}=-2dx\)

Integrate:

\(\displaystyle ln(Q)=-2x+c\)

\(\displaystyle Q=Ce^{-2x}\)

Now, use your initial condition to find C.

 

[Deleted member 4993]

Re: Differential Equation

I need a little help with dQ/dx = -2Q where Q = 40 when x = 0

I don't think I am doing it correctly. Since I don't know the answer I have no idea if I am on the right track.

A)

dQ/dx = -2Q

This is a seperable variable problem - your textbook should have many examples of this type.

As an example I'll do a different (but somewhat similar) problem

dG/dx = G^2 with G = 5 at x = 0

\(\displaystyle \frac{dG}{dx} \, = \,G^2\)

\(\displaystyle \frac{dG}{G^2} \, = \,dx\)

\(\displaystyle \int\frac{dG}{G^2} \, = \,\int dx\)

\(\displaystyle -\frac{1}{G} \, = \,x\, + \, C\)

Now apply initial condition

G = 5 for x = 0

then

-1/5 = 0 + C

C = -1/5

then

\(\displaystyle G(x)\, = \, \frac{5}{1\, - \, 5\cdot x}\)

Now to check correctness of your answer

you should apply initial condition( x= 0) above and see what do you get for G

and

differentiate the equation above and see if you get back your original DE.


Follow the exact same procedure for your given problem

y = dQ = -2Q dx and x = q

= -2[q^1/1] + 1[q^2/2] + C

= -2q + q^2 + C

B)

40 = -2(0) + (0)^2 + C

40 = C

Q = -2q + q^2 + 40


I understand that once you find the answer formula that you set it equal to 40 and plug 0 in for each x. But I don't think that I have part A correct.