# Math Help - 2nd order DE constant coefficients and boundary conditions.

1. ## 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

2. What ideas have you had so far?

3. 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

4. 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?

5. 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

$m=\alpha\pm i\beta$

$v=e^{x\alpha}(C_1\cos{\beta}+C_2i\sin{\beta})$

6. 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?

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

8. 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
$\displaystyle m^2-4m+5=0\Rightarrow m=\frac{4\pm\sqrt{16-4(5)}}{2}=\frac{4\pm 2i}{2}=2\pm i$

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

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

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

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

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

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

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