# Integrating IVP

• Nov 3rd 2009, 11:49 AM
matlabnoob
Integrating IVP

dx/dt = (x^3).(e^t)

intial condition : x(1) = -4

this looks v.simple but trust me, its not!

:( help me out! aaaah...

thank you
• Nov 3rd 2009, 11:53 AM
artvandalay11
Quote:

Originally Posted by matlabnoob

dx/dt = (x^3).(e^t)

intial condition : x(1) = -4

this looks v.simple but trust me, its not!

:( help me out! aaaah...

thank you

$\displaystyle \frac{dx}{dt}=x^3e^t$

$\displaystyle \frac{dx}{x^3}=e^tdt$

Now integrate both sides

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

can you handle it from here by plugging in your intial value to find c
• Nov 3rd 2009, 11:56 AM
matlabnoob
Quote:

Originally Posted by artvandalay11
$\displaystyle \frac{dx}{dt}=x^3e^t$

$\displaystyle \frac{dx}{x^3}=e^tdt$

Now integrate both sides

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

can you handle it from here by plugging in your intial value to find c

thank you!

i have gotten that far and then i ended up with =S...
after i substituted 1 into t..

x^2 = -1/2e+2c

?

that looks v.v.wrong to me. or is it just me =S
• Nov 3rd 2009, 12:01 PM
artvandalay11
Quote:

Originally Posted by matlabnoob
thank you!

i have gotten that far and then i ended up with =S...
after i substituted 1 into t..

x^2 = -1/2e+2c

?

that looks v.v.wrong to me. or is it just me =S

$\displaystyle x(1)=-4$ so when t=1, x=-4

$\displaystyle -\frac{1}{2(-4)^2}=e^1+c$

$\displaystyle -\frac{1}{32}=e+c$

So $\displaystyle c=-e-\frac{1}{32}$

And the solution is

$\displaystyle -\frac{1}{2x^2}=e^t-e-\frac{1}{32}$

which can be combined or manipulated or whatever you wanna do but I'll leave it like that
• Nov 3rd 2009, 12:07 PM
matlabnoob
thank you!=] woaah its that easy.... i spent 2 hourson this!!