Some help with this problem.Find the solution of the equation

v''- 4v'+5v=0,such that v=-1 and v'=1 when x=pi=3.14159

Thank you for your time.

Printable View

- Feb 1st 2011, 11:41 AMAkilMAI2nd order DE constant coefficients and boundary conditions.
Some help with this problem.Find the solution of the equation

v''- 4v'+5v=0,such that v=-1 and v'=1 when x=pi=3.14159

Thank you for your time. - Feb 1st 2011, 11:42 AMAckbeet
What ideas have you had so far?

- Feb 1st 2011, 11:53 AMdwsmith
- Feb 1st 2011, 12:13 PMAkilMAI
and then when I get m I can use the initial conditions to find the constant c1 and c2(for example) from the general solution form?

- Feb 1st 2011, 12:16 PMdwsmith
- Feb 5th 2011, 05:11 AMAkilMAI
now I get this system

e^[pi+2](A*cos[2x]+B*i*sin[2x])=-1

2e^[pi+2](B*i*cos[2x]-A*sin[2x])=1...how can I solve it? - Feb 5th 2011, 05:19 AMAkilMAI
sorry about that...

e^[pi+2](A*cos[2pi]+B*i*sin[2x])=-1=>e^[pi+2]*B*i=-1

2e^[pi+2](B*i*cos[2pi]-A*sin[2x])=1.=>2e^[pi+2]*A=1 - Feb 5th 2011, 11:26 AMdwsmith
$\displaystyle \displaystyle m^2-4m+5=0\Rightarrow m=\frac{4\pm\sqrt{16-4(5)}}{2}=\frac{4\pm 2i}{2}=2\pm i$

$\displaystyle \alpha=2, \ \ \ \beta=1$

$\displaystyle v=e^{2x}\left(C_1\cos(x)+C_2\sin(x)\right)$

$\displaystyle v(\pi)=e^{2\pi}\left(C_1\cos(\pi)+C_2\sin(\pi)\rig ht)=-1\Rightarrow e^{2\pi}\left(-1\cdot C_1)=-1$

$\displaystyle \displaystyle C_1e^{2\pi}=1\Rightarrow C_1=e^{-2\pi}$

$\displaystyle v'=e^{2x-2\pi}\left((C_2e^{-2\pi}+2)\cos(x)+(2e^{2\pi}C_2-1)\sin(x)\right)$

$\displaystyle \displaystyle v'(\pi)=(C_2e^{-2\pi}+2)\cos(\pi)+(2e^{2\pi}C_2-1)\sin(\pi)=1\Rightarrow -(C_2e^{-2\pi}+2)=1\Rightarrow C_2=-3e^{2\pi}$

$\displaystyle v(x)=e^{2x-2\pi}(\cos(x)-3e^{4\pi}\sin(x))$