SEcond order non-homogeneous eqn

[KLS2111]

Hello,
I am working on solving a second order non-homogeneous eqn.

The equation is d^2 x + 3 dx - 4x = tan(t)
dt^2 dt with x(0)=2 x'(0)=-1 Find x(1)

I found x to be 1/5 [ e^(-t) Integral e^ (t) tan t dt - e^(-4t) integral e^(4t) tan t dt

The issue I'm having is actually integrating this to plug in zero. When I integrate i ended up with
1/5[ e^t ln|sec t| - int ln |sec t| dt - e^4t ln|sect|- integ ln|sect| dt

Any guidance would be greatly appreciated

 

[galactus]

Is this it?:

$\displaystyle x''+3x'-4x=tan(t), \;\ x'(0)=-1, \;\ x(0)=2$

$\displaystyle \text{Find x(1)}$.

From $\displaystyle m^{2}+3m-4=(m+4)(m-1)=0$

$\displaystyle m=-4, \;\ m=1$

$\displaystyle y_{c}=C_{1}e^{-4t}+C_{2}e^{t}$

Using Variation of Parameters:

$\displaystyle W=4t$

$\displaystyle W_{1}=-t\cdot tan(t)$

$\displaystyle W_{2}=-4t\cdot tan(t)$

$\displaystyle u'_{1}=\frac{W_{1}}{W}=\frac{-tan(t)}{4}$

$\displaystyle u'_{2}=-tan(t)$

Integrating gives:

$\displaystyle u_{1}=\frac{ln(cos(t))}{4}$

$\displaystyle u_{2}=ln(cos(t))$

$\displaystyle \frac{e^{-4t}ln(cos(t))}{4}+e^{t}ln(cos(t))$

Using the initial conditions, we get a general solution $\displaystyle y_{c}+y_{p}=y$

$\displaystyle y=\frac{3}{5}e^{-4t}+\frac{7}{5}e^{t}+\frac{e^{-4t}ln(cos(t))}{4}+e^{t}ln(cos(t))$

It would appear you are trying to integrate what was already integrated. It is tough to integrate ln(cos(t)).

Actually, this is the result after integrating.

Let t=1 to find y(1).

 

[KLS2111]

Yes that is the eqn thanks