y"-3y'+2y=0 y=0 and y'=2 when x=0

Question

y"-3y'+2y=0  y=0 and y'=2 when x=0

anonymous · Answer

What are the instructions it gives you?

anonymous · Answer

find the particular solution of the differential equation that satisfies the stated conditions...it's a exercise of the SWOKOWSKI book

loser66 · Answer

it's a homogeneous , how to have particular solution which is just for non-homogenous form?

loser66 · Answer

you have general solution only.

whpalmer4 · Answer

Okay, assume you'll have a solution of the form $y = e^{\lambda x}$

$$y'(x) = \lambda e^{\lambda x}$$$$y''(x) = \lambda^2 e^{\lambda x}$$

$$y''(x) - 3y'(x) +2y(x) = 0$$$$\lambda^2 e^{\lambda x} - 3\lambda e^{\lambda x} + 2e^{\lambda x} = 0$$factoring you have $$e^{\lambda x}(\lambda^2 - 3\lambda + 2) = 0$$but $e^{\lambda x} = y(x)$ so $$y(x)(\lambda^2 - 3\lambda + 2) = 0$$solutions to that are where $y(x) = 0$ (nowhere) or roots of the quadratic, $\lambda = 1, \lambda = 2$

So your general solution is $y(x) = c_1 e^{\lambda x} + c_2 e^{\lambda x}$ or$$y(x) = c_1e^{x} + c_2e^{2x}$$

Plug in your initial conditions to solve for the values of $c_1, c_2$

whpalmer4 · Answer

$$y(0) = c_1e^{0} + c_2e^{2*0}$$$$y'(x) = c_1 e^x + 2c_2 e^{2x}$$$$y'(0) = 2 = c_1e^{0} + 2c_1 e^{2*0}$$

That gives you enough information to solve for $c_1,c_2$

whpalmer4 · Answer

isn't differentiation of $e^{ax}$ great? :-)

loser66 · Answer

oh, you don't know how to find characteristic equation, ok
 from the problem y'' stands for r^2
                           3y' stands for 3r
                           2y stands for 2
so, the characteristic equation is r^2 -3r +2 =0 
solve for r , they are r1 = 1 and r2 =2
 got me?

loser66 · Answer

so that the general solution will have the form of $C_1 e^{r_1x} + C_2e^{r_2x}$ which is $y = C_1e^x +C_2e^{2x}$ so far not good???? hahahaha.... I am waiting for your confirm whether you got me or not

loser66 · Answer

Ok, good. the sentence y =0, y' =2 when x =0 means 
y (0) = 0
y'(0) =2.
Dat sit
now, apply on our solution, we have y =..... that is the short form of y(x) = $C_1e^x + C_2e^{2x}$
 so, y (0)? just plug 0 into the solution where you see x
so that y (0) =$C_1 e^0 +C_2e^{2*0}= 0$
so, you have $C_1 + C_2 =0 (let say (1)$)
 take the derivative of the solution to get y'(x) = ...?
and then plug 2 into it to get y'(2) =.... you will have another equation let say (2)
solve (1) and (2) to get C1, C2
Plug them back to y (x) 
Done