# Thread: Simple Initial Value Problem..

1. ## Simple Initial Value Problem..

hi guys,
ok, so im reading my lecture notes again and here goes..

Solve the IVP dx/dt = 2x^2 with condition x(5) = -3

Solution: x(t) = -1/c + 2t
At t=5, we require x(5) = -3 (not equal to 0). We can use the solution x(t) = -1/c+2t, and determine c from the data given, i.e. -3 = -1/c+2t
so c = -29/3

now this is where i dont understand! (how did they get 6t ?? )

and hence the solution on the IVP is given by x(t) = 3/(29 - 6t)

thanks guys!!

2. Originally Posted by matlabnoob
hi guys,
ok, so im reading my lecture notes again and here goes..

Solve the IVP dx/dt = 2x^2 with condition x(5) = -3

Solution: x(t) = -1/c + 2t
At t=5, we require x(5) = -3 (not equal to 0). We can use the solution x(t) = -1/c+2t, and determine c from the data given, i.e. -3 = -1/c+2t
so c = -29/3

now this is where i dont understand! (how did they get 6t ?? )

and hence the solution on the IVP is given by x(t) = 3/(29 - 6t)

thanks guys!!
I think there's a problem. . When you spearate

$\displaystyle \frac{dx}{x^2} = 2 dt$ so $\displaystyle - \frac{1}{x} = 2t + c$ or $\displaystyle x = \frac{-1}{2t+c}$, not what you have.

3. aah of course ,how did i miss that?

so its -1/c'+2t..

im trying to get 6t somehow as the result does... is it possible to get 6t? =S because what im doing is plugging in c = -29/3 into x(t) and t=5 into x(t)... but im not getting any 6ts..

4. Originally Posted by Danny
I think there's a problem. . When you spearate

$\displaystyle \frac{dx}{x^2} = 2 dt$ so $\displaystyle - \frac{1}{x} = 2t + c$ or $\displaystyle x = \frac{-1}{2t+c}$, not what you have.
Now use the IC $\displaystyle x(5) = -3$ so $\displaystyle -3 = \frac{-1}{10+c}$ giving $\displaystyle c = - \frac{29}{3}.$

Therefore $\displaystyle x = \frac{-1}{2t - \frac{29}{3}} = \frac{3}{29-6t}$.

5. Originally Posted by matlabnoob
hi guys,
ok, so im reading my lecture notes again and here goes..

Solve the IVP dx/dt = 2x^2 with condition x(5) = -3

Solution: x(t) = -1/c + 2t
At t=5, we require x(5) = -3 (not equal to 0). We can use the solution x(t) = -1/c+2t, and determine c from the data given, i.e. -3 = -1/c+2t
so c = -29/3

now this is where i dont understand! (how did they get 6t ?? )

and hence the solution on the IVP is given by x(t) = 3/(29 - 6t)

thanks guys!!
Alternative method:

$\displaystyle \frac{dx}{dt} = 2x^2$

$\displaystyle \frac{dt}{dx} = \frac{1}{2x^2}$

$\displaystyle \frac{dt}{dx} = \frac{1}{2}x^{-2}$

$\displaystyle t = \int{\frac{1}{2}x^{-2}\,dx}$

$\displaystyle t = -\frac{1}{2}x^{-1} + c$

$\displaystyle \frac{1}{2x} = c - t$

$\displaystyle 2x = \frac{1}{c - t}$

$\displaystyle x = \frac{1}{2c - 2t}$.

$\displaystyle x = \frac{1}{C - 2t}$, where $\displaystyle C = 2c$.

Now by using the fact that $\displaystyle x(5) = -3$

$\displaystyle -3 = \frac{1}{C - 2(5)}$

$\displaystyle -3 = \frac{1}{C - 10}$

$\displaystyle -\frac{1}{3} = C - 10$

$\displaystyle C = \frac{29}{3}$.

Therefore

$\displaystyle x = \frac{1}{\frac{29}{3} - 2t}$

$\displaystyle x = \frac{3}{29 - 6t}$.