# 2nd order DE constant coefficients and boundary conditions.

• February 1st 2011, 11:41 AM
AkilMAI
2nd 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
• February 1st 2011, 11:42 AM
Ackbeet
What ideas have you had so far?
• February 1st 2011, 11:53 AM
dwsmith
Quote:

Originally Posted by AkilMAI
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

You can treat this as a polynomial.

$m^2-4m+5=0$

Solve for m
• February 1st 2011, 12:13 PM
AkilMAI
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?
• February 1st 2011, 12:16 PM
dwsmith
Quote:

Originally Posted by AkilMAI
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?

When you get m, post it.

Your solution will be of the form

$m=\alpha\pm i\beta$

$v=e^{x\alpha}(C_1\cos{\beta}+C_2i\sin{\beta})$
• February 5th 2011, 05:11 AM
AkilMAI
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?
• February 5th 2011, 05:19 AM
AkilMAI
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
• February 5th 2011, 11:26 AM
dwsmith
Quote:

Originally Posted by AkilMAI
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

$\displaystyle m^2-4m+5=0\Rightarrow m=\frac{4\pm\sqrt{16-4(5)}}{2}=\frac{4\pm 2i}{2}=2\pm i$

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

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

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

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

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

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

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