IVP -variation of parameters

matlabnoob · #1 $Old$ November 3rd, 2009, 02:19 PM

ok so here is my *big* problem im stuck on..=[

thank you for your time reading this

dx/dt + x/t - 1 = t^2 ; x(1) = 1

using variation of paramter i did all this:

dx/dt + x/t = t^2 + 1

dx/dt+ x/t = 0 [solve homogeneous equation first]

integrating by parts i get. ..

-lnx = -lnt + c

x = e^(lnt+c) = At

then i find x(t), x'(t)...

x(t) = At, x'(t) = A

now what i did is put these into the ode which was dx/dt + x/t - 1 = t^2

and i get..

A + At/t = t^2 + 1

2A = t^2 + 1

A = (t^2 + 1)/2

and put A into x(t)...

i get..

x = ((t^2+1)/2)*t

and x(1) = 1.... now im stuck!

please could anyone guide me through or tell me what i did wrong? =[ am having a nightmare with these

Danny · #2 $Old$ November 3rd, 2009, 03:04 PM

Quote:

Originally Posted by matlabnoob $View Post$

ok so here is my *big* problem im stuck on..=[

thank you for your time reading this

dx/dt + x/t - 1 = t^2 ; x(1) = 1

using variation of paramter i did all this:

dx/dt + x/t = t^2 + 1

dx/dt+ x/t = 0 [solve homogeneous equation first]

integrating by parts i get. ..

-lnx = -lnt + c

x = e^(lnt+c) = At

then i find x(t), x'(t)...

x(t) = At, x'(t) = A

now what i did is put these into the ode which was dx/dt + x/t - 1 = t^2

and i get..

A + At/t = t^2 + 1

2A = t^2 + 1

A = (t^2 + 1)/2

and put A into x(t)...

i get..

x = ((t^2+1)/2)*t

and x(1) = 1.... now im stuck!

please could anyone guide me through or tell me what i did wrong? =[ am having a nightmare with these

First off, the solution $x = At$ doesn't solve the complementary ODE $x' + \frac{x}{t} = 0$ (it's $x = \frac{A}{t}$ ). Second, for a variation of parameters let $A = A(t),$ sub into the ODE and then reduce to an ODE for $A(t).$