Math Help - Finding the complementary equation of ...

1. Finding the complementary equation of ...

L[y]=cy''+(2-c)y'-y=e^c

I'm not sure how to approach this equation. I've already found the particular solution for this, which is Y(t)=e^c

I tried to find the complementary solution for it by using the quadratic equation and trying to find the required numbers that way, but I've been told that that method doesn't work. But here's the work i've done so far:

Work: tu^2+(2-t)u-1=0
tu^2-tu+2u-1=0

(let t=c)

when t=/0, u=+-(an ugly imaginary no.)

when t=0, 2u-1=0
u=1/2

thus, com. function is y=c1e^1/2t+c2e^1/2t

I've been told that this is wrong. I'm still really confused as to what to do...

EDIT: e^c not e^t

2. Originally Posted by TheFirstOrder
cy''+(2-c)y'-y=e^t

I'm not sure how to approach this equation. I've already found the particular solution for this, which is Y(t)=e^t.

I tried to find the complementary solution for it by using the quadratic equation and trying to find the required numbers that way, but I've been told that that method doesn't work. But here's the work i've done so far:

Work: tu^2+(2-t)u-1=0
tu^2-tu+2u-1=0

when t=/0, u=+-(an ugly imaginary no.)

when t=0, 2u-1=0
u=1/2

thus, com. function is y=c1e^1/2t+c2e^1/2t

I've been told that this is wrong. I'm still really confused as to what to do...
Dear TheFirstOrder,

$cy''+(2-c)y'-y=e^t$

As you have found out $e^{t}$ is a particular integral.

Also, $u=\frac{\sqrt{(-c+2)^2+4 c}+c-2}{2 c}~or~u=\frac{-\sqrt{(-c+2)^2+4 c}+c-2}{2 c}$ ; where u is the complimentary function.

Therefore, the general solution is,

$y=e^{y}+Ae^{\frac{\sqrt{(-c+2)^2+4 c}+c-2}{2 c}}+Be^{\frac{-\sqrt{(-c+2)^2+4 c}+c-2)}{2 c}}$ ; where A and B are arbitary constants.

3. [quote=Sudharaka;481123]Dear TheFirstOrder,

$cy''+(2-c)y'-y=e^t$

As you have found out $e^{t}$ is a particular integral.

Also, $u=\frac{\sqrt{(-c+2)^2+4 c}+c-2}{2 c}~or~u=\frac{-\sqrt{(-c+2)^2+4 c}+c-2}{2 c}$ ; where u is the complimentary function.

Therefore, the general solution is,

$y=e^{y}+Ae^{\frac{\sqrt{(-c+2)^2+4 c}+c-2}{2 c}}+Be^{\frac{-\sqrt{(-c+2)^2+4 c}+c-2)}{2 c}}$ ; where A and B are arbitary constants.

But can we use this method? (the coefficients are not constant...)

4. Originally Posted by TheFirstOrder
cy''+(2-c)y'-y=e^t

I'm not sure how to approach this equation. I've already found the particular solution for this, which is Y(t)=e^t.

I tried to find the complementary solution for it by using the quadratic equation and trying to find the required numbers that way, but I've been told that that method doesn't work. But here's the work i've done so far:

Work: tu^2+(2-t)u-1=0
tu^2-tu+2u-1=0

when t=/0, u=+-(an ugly imaginary no.)

when t=0, 2u-1=0
u=1/2

thus, com. function is y=c1e^1/2t+c2e^1/2t

I've been told that this is wrong. I'm still really confused as to what to do...
Is $c$ a constant, or a function of $x$ or $y$?

If it is a constant, you can use this method.

If not, you will need another way.