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